Calculus
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DIFFERENTIATION - I Let us take the graph of a function f . We know that the point ( x,f(x) ) is a point on the graph. What line, if any, should be the tangent to the graph at this point?
We can take another point on ( x+h, f(x+h) ) on the graph and draw a secant line through these two points. Keeping ( x,f(x) ) fixed, we move the other point closer to it (h tends to zero from the left or from the right). The secant line tends to a limiting position. This is “the tangent to the graph at the point ( x,f(x) )”. If this limit exists, that is defined to be the slope of the tangent line at that fixed point ( x,f(x) ) on the graph of y = f (x).
DIFFERENTIABILITY & THE DERIVATIVE OF A FUNCTION Differentiation is the process of finding the derivative of a function. A function f is said to be differentiable at x if and only if ff xf + hf @ ff xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim exists. hQ0 h `
a
` a
` a
If this limit exists, it is called the derivative of f at x and is denoted by f. x . ff xf +f h @ ff xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Therefore, f. x = lim . hQ 0 h ` a
`
a
` a
` a
If we draw a tangent to the graph at this point, the slope of the tangent is equal to f. x .
The derivative of a function y = f(x) with respect to x at a point x = x 0 is given by ff xf +f hf @ ff xf 0f 0f ` a f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f. x 0 = lim f provided that this limit exists. hQ0 h b
c
`
a
This limit is also called the instantaneous rate of change of y with respect to x at x = x 0 . ` a f df f f f f f f ` a dy f f f f f f f f x = f The d/dx notation : If y = f(x), then we can write f. x = dx dx
The derivative of the function y =f(x) with respect to x may be indicated by any one of ` a dy f f f f f f f f y , , y. , f. x , Dx , Dx f, the symbols dx dx ` a df f f f f f f f
` a
df xf f f f f f f f f f f f f f f f f . We can use any of these, dx
while solving problems.
•
A function is said to be differentiable at a point x = x 0 if the derivative of the function exists at that point.
•
Also, if a function is differentiable at every point in an interval, the function is said to be differentiable on that interval.
•
If we say “a function (of x) is differentiable” without mentioning the interval, we mean that it is differentiable for every value of x in its domain.
Example : 1. The function
f x = x 3 @ 3x ` a
is differentiable on the interval (-1,1) as it is
differentiable at every point in that interval. 2. The function f x = |x @ 2| is not differentiable on the interval (1,3) as it is not ` a
differentiable at x = 2 which lies in that interval. Example : Find f’(x) for the function f x = x 2 + 4 . ` a
Solution :
We formthe difference quotient B` C B C a2 ` a ` a 2 x + h + 4 @ x + 4 @ ff xf ff xf + hf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f
= h h 2 2 2 2 +f 2xh +f hf +f 4f @ xf @ 4f xf +f hf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f 2xh f f f f f f f f f f f f f f f f f f f f f f f f f = = = 2x + h h h
ff xf + hf @ ff xf ` a f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Therefore f. x = lim f = lim 2x + h = 2x hQ0 h Q 0 h `
` a
a
` a
Example : Find f. @ 3 and f. @ 1 given that f x = x 2 Solution :
`
a
`
a
` a
ff @ 3f + hf @ ff @ 3f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f By definition f. @ 3 = lim f hQ0 h ` a2 ` a2 @ 3f + hf @ @ 3f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h c b 2 9f @ 6h +f hf @ 9f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h 2 @f 6h +f hf f f f f f f f f f f f f f f f f f f f f f f f f f f = lim f hQ0 h ` a = lim @ 6 + h `
`
a
hQ0
=@6
Similarly
a
`
a
ff @ 1f + hf @ ff @ 3f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f. @ 1 = lim f hQ0 h ` a2 ` a2 @ 1f +f hf @ @ 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h ` a = lim @ 2 + h `
`
a
a
`
a
hQ0
=@2
` a
` a
Alternative method : Find f. x from f(x) first, and to put x=-3 and x=-1 in f. x to get `
a
`
a
f. @ 3 and f. @ 1 as shown below ff xf + hf @ ff xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f. x = lim hQ0 h ` a2 2 xf + hf @ xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h b c 2 2 2 x + 2xh + h @ xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h 2 2xh +f hf f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h ` a = lim 2x + h `
` a
a
` a
hQ0
= 2x
Therefore f. @ 3 = 2 A @ 3 = @ 6 ` a ` a and f. @ 1 = 2 A @ 1 = @ 2 `
a
`
a
Example : Find the derivative of f x = x 3 + 5 at the point (2,13). ` a
Solution :
f x = x3 + 5 ` a
` a
f. x
= = = = =
ff xf + hf @ ff xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim f hQ0 h B` C B C a3 3 x + h + 5 @ x + 5 f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim f hQ0 h b c 2 3 3 2 3 x + 3x h + 3xh + h + 5 @ xf @ 5f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim hQ0 h 2 3 2 3x hf +f 3xh +f hf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim f hQ0 h b c 2 lim 3x 2 + 3xh + h `
a
` a
hQ0 2
= 3x
Therefore f. 2 = 3.2 ` a
2
= 12
Hence the derivative of f x = x 3 + 5 at the point 2,13 is 12 A ` a
b
c
COMPARISON OF DIFFERENTIABILITY AND CONTINUITY If a function f is differentiable x, it is continuous at x.
But if a function is continuous at x, it may or may not be differentiable at x. Example : 1. The function f x = x 3 @ 3x is differentiable on its whole domain (which is R). ` a
As differentiability implies continuity, it is continuous at every point. 2. f(x) = |x| is continuous at any real value of x. But as continuity does not imply differentiability, we have to check its differentiability at every at every point. We find that at x=0, it is not differentiable though continuous. 3. The function
2 ` a xf @ 4f f f f f f f f f f f f f f f f f f x = f is not continuous at x @2
x = 2 . As continuity is
necessary for differentiability, it is not differentiable at that point.
IMPLICIT DIFFERENTIATION dy f f f f f f f If y is a function of x that satisfies the equation x 3 y @ 4y + 2 = 0 and we are to find f , dx 2f f f f f f f f f f f f f f f f f f f and then differentiate it to get one way is to first solve it for y and write y = f 4 @ x3 2 b c@ 1 c@ 2 b c ` ab dy df df 6x 2f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f 3 3 2 = = 2 4 @ x = 2 @ 1 4 @ x @ 3x = b c2 . dx dx dx 4 @ x 3 4 @ x3
f
g
But had the equation been x 3 y @ 4 y 3 + 2 = 0 , we could not solve it for y. In that case dy f f f f f f f how would we get the expression for f ? dx Here, we use Implicit Differentiation method which mainly uses the chain rule. Till now dy dy f f f f f f f f f f f f f f we always got f in terms of x only. But in the cases like this, f involves y too apart dx dx from x. Writing the equation again, x 3 y @ 4 y 3 + 2 = 0 differentiating both sides fo the equation with respect to x, we get b c b c ` a f ` a df df df df f f f f f f f 3 f f f f f f 3 f f f f f f f f f f f x y @ f 4y + f 2 = f 0 dx dx dx dx b c b c b c F ` aG f df df df f f f f f f 3 f f f f f f f f f f f 3 dy f f f f f f f yA f x + x 3A f y @ f 4y A f + 0 = 0 using chain rule wherever required dx dx dy dx dy dy f f f f f f f f f f f f f f y A 3x 2 + x 3 A f @ 12 y 2 A f = 0 dx dx 2 dy 3x yf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = f 2 3 dx 12 y @ x
dy f f f f f f f is no disadvantage. This is quite common in practical problems. Many a Having y in f dx cases equation of the curve is such that the slope of the tangent on it at any point may depend on the y value at that point too. dy dx f f f f f f f f f f f f f f is easier than finding f . In that case, we can use the relation Sometimes, finding f dx dy
dy 1f f f f f f f f f f f f f f f f = dx f f f f f f f f f dx dy
dy w w w w w w w f f f f f f f given that x = p y @ 2 Example : Find f dx
Solution : First method :
w w w w w w w
x =py @2 w w w w w w w py =x + 2
y = x + 2 = x 2 + 4x + 4 dy f f f f f f f f = 2x + 4 dx Second method : w w w w w w w x = py @2 dx 1f f f f f f f f f f f f f f f f f f f f f f f f f = f w w w w w w w p 2 y dy dy w w w w w w w w w w w w w w f f f f f f f f = 2 p y putting p y = x + 2 we get dx = 2x + 4 `
a2
Example : Find the slope of the curve x 2 + y 2 = 25 at the point (3,4). Solution : x 2 + y 2 = 25 differentiating both sides with respect to x we get b c b c ` a df df df f f f f f f f 2 f f f f f f f 2 f f f f f f f x + y = 25 using chain rule we get dx dx dx b c df f f f f f f f 2 dy f f f f f f f f 2x + y A =0 dy dx dy f f f f f f f f 2x + 2y A =0 dx dy yf f f f f f f f f f f f f =@ dx x b c 4f f f f Slope of the curve at 3,4 = @ 3
We can take another point on ( x+h, f(x+h) ) on the graph and draw a secant line through these two points. Keeping ( x,f(x) ) fixed, we move the other point closer to it (h tends to zero from the left or from the right). The secant line tends to a limiting position. This is “the tangent to the graph at the point ( x,f(x) )”. If this limit exists, that is defined to be the slope of the tangent line at that fixed point ( x,f(x) ) on the graph of y = f (x).
DIFFERENTIABILITY & THE DERIVATIVE OF A FUNCTION Differentiation is the process of finding the derivative of a function. A function f is said to be differentiable at x if and only if ff xf + hf @ ff xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim exists. hQ0 h `
a
` a
` a
If this limit exists, it is called the derivative of f at x and is denoted by f. x . ff xf +f h @ ff xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Therefore, f. x = lim . hQ 0 h ` a
`
a
` a
` a
If we draw a tangent to the graph at this point, the slope of the tangent is equal to f. x .
The derivative of a function y = f(x) with respect to x at a point x = x 0 is given by ff xf +f hf @ ff xf 0f 0f ` a f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f. x 0 = lim f provided that this limit exists. hQ0 h b
c
`
a
This limit is also called the instantaneous rate of change of y with respect to x at x = x 0 . ` a f df f f f f f f ` a dy f f f f f f f f x = f The d/dx notation : If y = f(x), then we can write f. x = dx dx
The derivative of the function y =f(x) with respect to x may be indicated by any one of ` a dy f f f f f f f f y , , y. , f. x , Dx , Dx f, the symbols dx dx ` a df f f f f f f f
` a
df xf f f f f f f f f f f f f f f f f . We can use any of these, dx
while solving problems.
•
A function is said to be differentiable at a point x = x 0 if the derivative of the function exists at that point.
•
Also, if a function is differentiable at every point in an interval, the function is said to be differentiable on that interval.
•
If we say “a function (of x) is differentiable” without mentioning the interval, we mean that it is differentiable for every value of x in its domain.
Example : 1. The function
f x = x 3 @ 3x ` a
is differentiable on the interval (-1,1) as it is
differentiable at every point in that interval. 2. The function f x = |x @ 2| is not differentiable on the interval (1,3) as it is not ` a
differentiable at x = 2 which lies in that interval. Example : Find f’(x) for the function f x = x 2 + 4 . ` a
Solution :
We formthe difference quotient B` C B C a2 ` a ` a 2 x + h + 4 @ x + 4 @ ff xf ff xf + hf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f
= h h 2 2 2 2 +f 2xh +f hf +f 4f @ xf @ 4f xf +f hf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f 2xh f f f f f f f f f f f f f f f f f f f f f f f f f = = = 2x + h h h
ff xf + hf @ ff xf ` a f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f Therefore f. x = lim f = lim 2x + h = 2x hQ0 h Q 0 h `
` a
a
` a
Example : Find f. @ 3 and f. @ 1 given that f x = x 2 Solution :
`
a
`
a
` a
ff @ 3f + hf @ ff @ 3f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f By definition f. @ 3 = lim f hQ0 h ` a2 ` a2 @ 3f + hf @ @ 3f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h c b 2 9f @ 6h +f hf @ 9f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h 2 @f 6h +f hf f f f f f f f f f f f f f f f f f f f f f f f f f f = lim f hQ0 h ` a = lim @ 6 + h `
`
a
hQ0
=@6
Similarly
a
`
a
ff @ 1f + hf @ ff @ 3f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f. @ 1 = lim f hQ0 h ` a2 ` a2 @ 1f +f hf @ @ 1f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h ` a = lim @ 2 + h `
`
a
a
`
a
hQ0
=@2
` a
` a
Alternative method : Find f. x from f(x) first, and to put x=-3 and x=-1 in f. x to get `
a
`
a
f. @ 3 and f. @ 1 as shown below ff xf + hf @ ff xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f. x = lim hQ0 h ` a2 2 xf + hf @ xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h b c 2 2 2 x + 2xh + h @ xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h 2 2xh +f hf f f f f f f f f f f f f f f f f f f f f f f f f f = lim hQ0 h ` a = lim 2x + h `
` a
a
` a
hQ0
= 2x
Therefore f. @ 3 = 2 A @ 3 = @ 6 ` a ` a and f. @ 1 = 2 A @ 1 = @ 2 `
a
`
a
Example : Find the derivative of f x = x 3 + 5 at the point (2,13). ` a
Solution :
f x = x3 + 5 ` a
` a
f. x
= = = = =
ff xf + hf @ ff xf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim f hQ0 h B` C B C a3 3 x + h + 5 @ x + 5 f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim f hQ0 h b c 2 3 3 2 3 x + 3x h + 3xh + h + 5 @ xf @ 5f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim hQ0 h 2 3 2 3x hf +f 3xh +f hf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f lim f hQ0 h b c 2 lim 3x 2 + 3xh + h `
a
` a
hQ0 2
= 3x
Therefore f. 2 = 3.2 ` a
2
= 12
Hence the derivative of f x = x 3 + 5 at the point 2,13 is 12 A ` a
b
c
COMPARISON OF DIFFERENTIABILITY AND CONTINUITY If a function f is differentiable x, it is continuous at x.
But if a function is continuous at x, it may or may not be differentiable at x. Example : 1. The function f x = x 3 @ 3x is differentiable on its whole domain (which is R). ` a
As differentiability implies continuity, it is continuous at every point. 2. f(x) = |x| is continuous at any real value of x. But as continuity does not imply differentiability, we have to check its differentiability at every at every point. We find that at x=0, it is not differentiable though continuous. 3. The function
2 ` a xf @ 4f f f f f f f f f f f f f f f f f f x = f is not continuous at x @2
x = 2 . As continuity is
necessary for differentiability, it is not differentiable at that point.
IMPLICIT DIFFERENTIATION dy f f f f f f f If y is a function of x that satisfies the equation x 3 y @ 4y + 2 = 0 and we are to find f , dx 2f f f f f f f f f f f f f f f f f f f and then differentiate it to get one way is to first solve it for y and write y = f 4 @ x3 2 b c@ 1 c@ 2 b c ` ab dy df df 6x 2f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f 3 3 2 = = 2 4 @ x = 2 @ 1 4 @ x @ 3x = b c2 . dx dx dx 4 @ x 3 4 @ x3
f
g
But had the equation been x 3 y @ 4 y 3 + 2 = 0 , we could not solve it for y. In that case dy f f f f f f f how would we get the expression for f ? dx Here, we use Implicit Differentiation method which mainly uses the chain rule. Till now dy dy f f f f f f f f f f f f f f we always got f in terms of x only. But in the cases like this, f involves y too apart dx dx from x. Writing the equation again, x 3 y @ 4 y 3 + 2 = 0 differentiating both sides fo the equation with respect to x, we get b c b c ` a f ` a df df df df f f f f f f f 3 f f f f f f 3 f f f f f f f f f f f x y @ f 4y + f 2 = f 0 dx dx dx dx b c b c b c F ` aG f df df df f f f f f f 3 f f f f f f f f f f f 3 dy f f f f f f f yA f x + x 3A f y @ f 4y A f + 0 = 0 using chain rule wherever required dx dx dy dx dy dy f f f f f f f f f f f f f f y A 3x 2 + x 3 A f @ 12 y 2 A f = 0 dx dx 2 dy 3x yf f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f = f 2 3 dx 12 y @ x
dy f f f f f f f is no disadvantage. This is quite common in practical problems. Many a Having y in f dx cases equation of the curve is such that the slope of the tangent on it at any point may depend on the y value at that point too. dy dx f f f f f f f f f f f f f f is easier than finding f . In that case, we can use the relation Sometimes, finding f dx dy
dy 1f f f f f f f f f f f f f f f f = dx f f f f f f f f f dx dy
dy w w w w w w w f f f f f f f given that x = p y @ 2 Example : Find f dx
Solution : First method :
w w w w w w w
x =py @2 w w w w w w w py =x + 2
y = x + 2 = x 2 + 4x + 4 dy f f f f f f f f = 2x + 4 dx Second method : w w w w w w w x = py @2 dx 1f f f f f f f f f f f f f f f f f f f f f f f f f = f w w w w w w w p 2 y dy dy w w w w w w w w w w w w w w f f f f f f f f = 2 p y putting p y = x + 2 we get dx = 2x + 4 `
a2
Example : Find the slope of the curve x 2 + y 2 = 25 at the point (3,4). Solution : x 2 + y 2 = 25 differentiating both sides with respect to x we get b c b c ` a df df df f f f f f f f 2 f f f f f f f 2 f f f f f f f x + y = 25 using chain rule we get dx dx dx b c df f f f f f f f 2 dy f f f f f f f f 2x + y A =0 dy dx dy f f f f f f f f 2x + 2y A =0 dx dy yf f f f f f f f f f f f f =@ dx x b c 4f f f f Slope of the curve at 3,4 = @ 3
