Derivative and related calculus expressions

Key concept: This set contains several derivative-style prompts. The common thread is applying the chain rule, product rule, and implicit differentiation correctly before evaluating at the requested point.

Step by step

1) If $f(x)=f(g(x))$

This appears to be a chain-rule reminder. For a composition,

$\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

2) If $f(x)=2^{x^2+2\ln(x^2+x)}$, find $f'(1)$

Let

$u(x)=x^2+2\ln(x^2+x)$

Then

$f(x)=2^{u(x)}$

So

$f'(x)=2^{u(x)}\ln(2)\cdot u'(x)$

Now

$u'(x)=2x+2\cdot\frac{2x+1}{x^2+x}$

At $x=1$:

$u(1)=1+2\ln(2)$

$u'(1)=2+2\cdot\frac{3}{2}=5$

Thus

$f'(1)=2^{1+2\ln(2)}\ln(2)\cdot 5$

3) For $x^2-y^2=8$, evaluate $\frac{dy}{dx}$

Differentiate implicitly:

$2x-2y\frac{dy}{dx}=0$

So

$\frac{dy}{dx}=\frac{x}{y}$

At any given point on the curve, substitute the coordinates into $\frac{x}{y}$.

4) Slope of the tangent to the graph of $x\,f(x)$ at $x=1$

Differentiate using the product rule:

$\frac{d}{dx}[x f(x)] = f(x)+x f'(x)$

So the slope at $x=1$ is

$f(1)+f'(1)$

5) If $m(x)=\frac{3x^3-2}{x}$, find the instantaneous rate of change at $x=2$

Rewrite first:

$m(x)=3x^2-\frac{2}{x}$

Differentiate:

$m'(x)=6x+\frac{2}{x^2}$

Then

$m'(2)=12+\frac{2}{4}=12.5=\frac{25}{2}$

Final notes

Each item is solved by choosing the correct differentiation rule first, then substituting the requested point.

Pitfall alert

The most common mistakes here are: forgetting the chain rule for composed functions, missing the derivative of the exponent inside $2^{u(x)}$, and substituting the point before differentiating. For implicit differentiation, remember that $y$ depends on $x$.

Try different conditions

If the logarithmic exponent in item 2 were changed, only the inner derivative $u'(x)$ would change. If the curve in item 3 were $x^2+y^2=8$, the derivative would become $\frac{dy}{dx}=-\frac{x}{y}$ instead of $\frac{x}{y}$.

Further reading

chain rule, implicit differentiation, product rule