Exponential and Sinusoidal Derivatives Practice

These questions follow Derivatives of Exponential Functions and Derivatives of Sinusoidal Functions — the two families that model growth and repetition, alone and in combination. Radians throughout, always.

Exponential derivatives

1. Differentiate $f(x)=e^x$ and $g(x)=2^x$.
2. Differentiate $f(x)=e^{3x}$.
3. Evaluate $\frac{2^h-1}{h}$ for $h=0.1$, $0.01$, and $0.001$. What is the march approaching, and what does it verify?
4. A population is modelled by $N(t)=500e^{0.2t}$, with $t$ in years. Determine the rate of growth at $t=5$.

Answer 1

$f^{\prime }(x)=e^x$ — the function that is its own derivative. $g^{\prime }(x)=2^x\ln 2$ — same shape, stretched by the constant $\ln 2\approx 0.693$. Only base $e$ escapes the stretch factor.

Answer 2

Chain rule: the inside $3x$ reports its rate 3, so $f^{\prime }(x)=3e^{3x}$. At every point this function grows at three times its own height.

Answer 3

$h=0.1$: $\frac{2^{0.1}-1}{0.1}\approx 0.718$. $h=0.01$: $\approx 0.696$. $h=0.001$: $\approx 0.6934$. The march settles toward $\ln 2\approx 0.693$ — and this limit is exactly the stretch factor in $\frac{d}{dx}{2}^x=2^x\ln 2$, verified with nothing but a calculator.

Answer 4

$N^{\prime }(t)=500\times 0.2e^{0.2t}=100e^{0.2t}$. At $t=5$: $N^{\prime }(5)=100e^1\approx 272$ individuals per year. Note the signature of exponential growth: $N^{\prime }(t)=0.2N(t)$ — the rate is always 20% of the size.

Sinusoidal derivatives

5. Determine the slope of $f(x)=\sin x$ at $x=\frac{\pi }{3}$.
6. Determine the equation of the tangent line to $f(x)=\cos x$ at $x=\frac{\pi }{2}$.
7. Differentiate $f(x)=x\sin x$.
8. Express $\tan x$ as $\frac{\sin x}{\cos x}$ and differentiate it using the product and chain rules. Simplify.

Answer 5

$f^{\prime }(x)=\cos x$, so the slope is $\cos \frac{\pi }{3}=\frac{1}{2}$. Check against the graph: at $\frac{\pi }{3}$ sine is still climbing toward its crest at $\frac{\pi }{2}$, but flattening — a modest positive slope fits.

Answer 6

Point: $f\left(\frac{\pi }{2}\right)=0$. Slope: $f^{\prime }(x)=-\sin x$, so $f^{\prime }\left(\frac{\pi }{2}\right)=-1$. Tangent: $y=-\left(x-\frac{\pi }{2}\right)$. Cosine crosses its axis at full steepness, heading down — slope exactly $-1$.

Answer 7

Product rule: $f^{\prime }(x)=1\cdot \sin x+x\cos x=\sin x+x\cos x$. Both factors take their turn; neither $x$ nor $\sin x$ gets to hold still for free.

Answer 8

Write $f(x)=\sin x(\cos x{)}^{-1}$. Then $f^{\prime }(x)=\cos x(\cos x{)}^{-1}+\sin x\cdot (-1)(\cos x{)}^{-2}(-\sin x)=1+\frac{{\sin }^{2}x}{{\cos }^{2}x}$ Over a common denominator: $\frac{{\cos }^{2}x+{\sin }^{2}x}{{\cos }^{2}x}=\frac{1}{{\cos }^{2}x}$ — the Pythagorean identity collapsing the sum. The derivative of $\tan x$ is $\frac{1}{{\cos }^{2}x}$, always positive: tangent only ever climbs.

Rates in the wild

9. The height of a tide is modelled by $h(t)=3\sin \left(\frac{\pi t}{6}\right)+5$ metres, with $t$ in hours. Determine the rate of change of the height at $t=3$, and interpret the answer.

Answer 9

Chain rule: $h^{\prime }(t)=3\cos \left(\frac{\pi t}{6}\right)\cdot \frac{\pi }{6}=\frac{\pi }{2}\cos \left(\frac{\pi t}{6}\right)$. At $t=3$: $h^{\prime }(3)=\frac{\pi }{2}\cos \frac{\pi }{2}=0$ metres per hour. Interpret before moving on: $h(3)=3\sin \frac{\pi }{2}+5=8$, the model’s maximum — this is high tide, and the water is momentarily still. A rate of zero at the crest is the model agreeing with the ocean.