Nested derivatives and increasing intervals

(a) Suppose the following facts are known about the function gg and its derivative: g(5)=1g(5)=1, g′(5)=−3g'(5)=-3. If f(x)=tan⁡(eg(x))f(x)=\tan(e^{g(x)}), find the value of f′(5)f'(5). marks

(b) Find, by differentiation, the range of values of xx for which the function y=x22x−1y=\frac{x^2}{2x-1} increases as xx increases. marks

[2009/DHS/Prelim/I/4]

Total marks: 6

Solution

(a)

Chain rule

Given g(5)=1g(5)=1 and g′(5)=−3g'(5)=-3, and f(x)=tan⁡(eg(x))f(x)=\tan\left(e^{g(x)}\right), find f′(5)f'(5).

Explore this idea

Hint

  1. There are three layers: tan⁡\tan, then e(⋅)e^{(\cdot)}, then g(x)g(x).

  2. Differentiate from the outside in, multiplying the derivatives.

  3. Substitute x=5x=5 and use radians.

Answer · (a)

f′(5)=−3esec⁡2(e)≈−9.81f'(5)=-3e\sec^2(e)\approx-9.81 (3 s.f.).

Radians.

Explore the idea

Three layers, three factors

Step through the layers from the inside out. At x = 5 the inner value is 1.

Teal: g(5) = 1. No comparison overlay.

Start: g(5) = 1

The innermost layer: g(5)=1g(5)=1 and g′(5)=−3g'(5)=-3. This supplies the factor −3-3.

  • Running product so far
  • Value after this layer
Ideas to try
Which layer changes the sign? Which changes the size most?
(b)

Increasing functions

Find, by differentiation, the range of values of xx for which y=x22x−1y=\dfrac{x^2}{2x-1} increases as xx increases.

Explore this idea

Hint

  1. Differentiate with the quotient rule.

  2. yy is increasing where dydx>0\frac{dy}{dx}>0.

  3. Solve the inequality, noting the denominator is a square.

Answer · (b)

x<0x<0 or x>1x>1.

Explore the idea

Where does y = x²/(2x − 1) rise?

The upper graph is the curve with its tangent. The lower graph is the gradient. Rising stretches are shaded in both.

Position along the curvex = -1
Try an example
Test a critical position

Buttons use full-precision reviewed positions; displayed coordinates are rounded.

f(x)
-0.33333
Gradient f′(x)
0.44444
positive

For x<0 the gradient is positive: the curve rises.

Selected values are within the original drawing windows.

  • Y = x²/(2x − 1)
  • Tangent at your position
  • Gradient, and where it is positive
  • Join P to Q; the dashed legs show Δx and Δy
Key idea

yy increases exactly where the gradient graph is above the axis: x<0x<0 or x>1x>1.

Ideas to try
  1. Move across x=0x=0 and x=1x=1: what does the tangent do?
  2. What happens to the gradient near x=12x=\frac12?