Nested derivatives and increasing intervals · source reference

Printed question

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

(b) Find, by differentiation, the range of values of $x$ for which the function $y=\frac{x^2}{2x-1}$ increases as $x$ increases. [3]

[2009/DHS/Prelim/I/4]

Printed working

(a) $f(x)=\tan(e^{g(x)})$.

$$f'(x)=\sec^2(e^{g(x)})\times e^{g(x)}\times g'(x).$$

Thus $f'(5)=\sec^2(e^1)\times e^1\times(-3)=-9.81$ (3 s.f.).

(b) $y=\frac{x^2}{2x-1}$.

$$\frac{dy}{dx}=\frac{(2x-1)(2x)-x^2(2)}{(2x-1)^2}=\frac{2x^2-2x}{(2x-1)^2}.$$

Function is increasing $\Rightarrow\frac{dy}{dx}>0\Rightarrow\frac{2x(x-1)}{(2x-1)^2}>0$.

Therefore $2x(x-1)>0$, since $(2x-1)^2>0$ for all real $x\ne\frac12$. The numerator parabola has zeros $0,1$ and is positive outside those zeros, so $x<0$ or $x>1$.

Teacher notes

Apply Chain Rule: let $u=g(x)$ and $v=e^{g(x)}=e^u$.

$$f'(x)=\frac{d}{dv}(\tan v)\frac{dv}{dx}=\sec^2v\frac{d}{du}(e^u)\frac{du}{dx}=\sec^2(e^{g(x)})e^{g(x)}g'(x).$$

Differentiate from “outside to inside”: layer ① is $\tan$, layer ② is the exponential $e^{\cdot}$, layer ③ is $g(x)$. Differentiate ① first, then ② → ③.

Apply quotient rule in (b).

Answer key

(a) $-9.81$ (3 s.f.). (b) $x<0$ or $x>1$.

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