Derivative of \ln(5x)?

\frac1x, because \ln5x=\ln5+\ln x

Why must x be in radians when differentiating trigonometric functions?

The results \frac{\sin h}{h}\to1 and \frac{d}{dx}\sin x=\cos x hold only for radians.

Define f'(x) from first principles.

\lim{h\to0}\frac{f(x+h)-f(x)}{h}

Differentiating exponentials, logarithms and trigonometric functionsAQA A-Level Maths: Flashcards

Card 1 of 120 of 12 known

Question

$\frac{d}{dx}(e^{kx})$?

Tap or press Space to reveal

Tap card or press Space to flip

See all 12 cards

$\frac{d}{dx}(e^{kx})$ ?: $ke^{kx}$
$\frac{d}{dx}(a^{kx})$ ?: $k\ln a\cdot a^{kx}$
$\frac{d}{dx}(\ln x)$ ?: $\frac1x$ for $x>0$
$\frac{d}{dx}(\sin kx)$ ?: $k\cos kx$
$\frac{d}{dx}(\cos kx)$ ?: $-k\sin kx$
$\frac{d}{dx}(\tan kx)$ ?: $k\sec^2kx$
Derivative of $\ln(5x)$ ?: $\frac1x$ , because $\ln5x=\ln5+\ln x$
Why must $x$ be in radians when differentiating trigonometric functions?: The results $\frac{\sin h}{h}\to1$ and $\frac{d}{dx}\sin x=\cos x$ hold only for radians.
Limit of $\frac{\sin h}{h}$ as $h\to0$ ?: $1$ (radians)
Limit of $\frac{\cos h-1}{h}$ as $h\to0$ ?: $0$ (radians)
Write $2^{x}$ as a power of $e$ .: $e^{x\ln2}$
Define $f'(x)$ from first principles.: $\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$

Exam questions on Differentiating exponentials, logarithms and trigonometric functions

A curve has equation $y=e^{3x}-4x$ .
Find the exact $x$ -coordinate of the stationary point of the curve.2 marks
A curve has equation $y=\tan2x$ for $-\frac{\pi}{4}<x<\frac{\pi}{4}$ , where $x$ is in radians.
Find the equation of the tangent to the curve at $x=\frac{\pi}{8}$ .2 marks
The population $P$ of a colony of bacteria, in thousands, is modelled by $P=5\times2^{0.4t}$ , where $t$ is the time in hours after the colony is first observed.
Find $\frac{dP}{dt}$ in terms of $t$ .3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).