Differentiation Of Inverse Functions

A2: Differentiation of an Inverse Function e.g. The curve f(x) = x2 – 4x -1 can be written as f(x) = (x – 2)2 – 5 and is shown in the diagram with a copy of inverse function

(4,5)

f-1(x) = 2 + √𝑥 + 5 .

Calculate the gradient of the inverse function at the point (4, 5).

(-5,2)

We find the gradient of the original function at the point (where

(2, -5)

the coordinates are reflected/ swapped). Then we use the fact that

𝑑𝑥 𝑑𝑦

1

= 𝑑𝑦

⁄𝑑𝑥

y = x2 – 4x – 1 differentiates to 5 (not x = 4) …

𝑑𝑦 𝑑𝑥

𝑑𝑦 𝑑𝑥

= 2𝑥 − 4. We now substitute x =

= 2 × 5 − 4 = 10 − 4 = 6 1

 Required gradient = 6.

Exercise: 1. a. Sketch the quadratic function y = 2x2 – 12x +7 for the domain x ≥ 3. (You may complete the square) b. Add the sketch of the inverse function and the point A(-3, 5). c. Find, by differentiation, the gradient of the inverse function at point A. 2. A function f(x) = x2(4 – x) is defined over the domain 0 ≤ x ≤ 8⁄3 and has an inverse function of f-1(x) (1-1 mapping). Find the gradient of f-1(x) at the point (7⁄8 , 1⁄2) . 3. a. Given that g(x) = x2(x – 2), for x > 2, calculate the gradient of the graph of y = g(x) at the point (4, 32). b. Deduce the gradient of the graph y = g-1(x) at the point (32, 4). 4. The function y = ln(1 + 2x) is defined by the domain x ≥ 0. By differentiating, or otherwise, find the gradient of the inverse function at the point (ln5, 2) 𝑒𝑥

5. The function f(x) = 2−𝑒 𝑥 can be represented by a 1 – 1 mapping

when its domain is restricted 0 < x < ln2. Show that the gradient of the inverse function at the point (1, 0) is ½.

6. The function y = 2 + ½sin(2x) completes one cycle in the domain 𝜋

𝜋

− 2 ≤ 𝑥 ≤ 2. Find the gradient of the inverse function at the point (2, 0).

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A2: Differentiation of an Inverse Function 1. Completing the square y = 2[x2 – 6x +

7 2

4. y = ln(1 + 2x)

]

𝑑𝑦 1 = × (2) 𝑑𝑥 1 + 2𝑥

A(-3, 5)

= 2[(x – 3)2 – 9 + 7 2

]

(ln5, 2)

= 2(x – 3)2 - 11 (b)

𝑑𝑦 𝑑𝑥

= 4𝑥 − 12 2

When x = 5, gradient = 4 × 5 – 12 = 8

2

When x = 2, gradient = 1+2×2 = 5 1

Gradient of inverse function at (-3, 5) = 8

Gradient of inverse function at (ln5, 2) = 5 2

5. y = 𝑒 𝑥 2−𝑒 𝑥

2. y = x2(4 – x) = 4x2 – x3 𝑑𝑦 = 8𝑥 − 3𝑥 2 𝑑𝑥

(2 − 𝑒 𝑥 )[𝑒 𝑥 ] − 𝑒 𝑥 [−𝑒 𝑥 ] 𝑑𝑦 = (2 − 𝑒 𝑥 )2 𝑑𝑥 8

/3

4

=

2

When x = 2, gradient = (2− 𝑒 0 )2 = 12 = 2

13

1

4

Gradient of inverse function at (1, 0) = 2

4

Gradient of inverse function at (⅞, ½) = 13 3. y = x2(x – 2) = x3 – 2x2 𝑑𝑦 = 3𝑥 2 − 4𝑥 𝑑𝑥

(2−𝑒 𝑥 )2

2𝑒 0

When x = ½, gradient = 8 × ½ – 3×(½)2 =4-¾ = 3¼ =

2𝑒 𝑥

6. y = 2 + ½sin(2x) 4

/3

2

𝑑𝑦 1 = 𝑐𝑜𝑠(2𝑥) × 2 𝑑𝑥 2 = cos(2x)

When x = 4, gradient = 3×(4)2 - 4×4 = 48 – 16 = 32

When x = 0, gradient = cos(0) = 1 1

Gradient of inverse function at (32, 4) = 32

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Gradient of inverse function at (2, 0) = 1

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