Differentiation practice questions — IB Maths AA

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{(1\mathrm{/}x)\cdot x-\ln x\cdot 1}{x^2}}={\displaystyle \frac{1-\ln x}{x^2}}$. At $x=1$: ${\displaystyle \frac{1-0}{1}}=1$.

Outer: ${\sec }^2(x^2)$. Inner derivative: $2x$. So $f^{\mathrm{\prime }}(x)=2x{\sec }^2(x^2)$.

Apply the product rule with $u=(2x-1{)}^4$ and $v=x+1$.

Differentiate each part (chain rule on $u$): $$u^{\mathrm{\prime }}=4(2x-1{)}^3\cdot 2=8(2x-1{)}^3,v^{\mathrm{\prime }}=1.$$

So $$f^{\mathrm{\prime }}(x)=8(2x-1{)}^3(x+1)+(2x-1{)}^4\mathrm{.}$$

Factor out $(2x-1{)}^3$: $$f^{\mathrm{\prime }}(x)=(2x-1{)}^3[8(x+1)+(2x-1)]=(2x-1{)}^3(10x+7)\mathrm{.}$$

Substitute $x=1$, where $2x-1=1$: $$f^{\mathrm{\prime }}(1)=(1{)}^3(10+7)=17.$$

$f^{\mathrm{\prime }}(x)=2x-6$.

Setting $f^{\mathrm{\prime }}(c)=0$: $2c-6=0⟹c=3$.

Since $f^{\mathrm{\prime }\mathrm{\prime }}(x)=2>0$, this is indeed a minimum.

$f^{\mathrm{\prime }}(x)=(x{)}^{\mathrm{\prime }}{e}^x+x(e^x{)}^{\mathrm{\prime }}=e^x+x{e}^x=e^x(x+1)$.

$f$ is a polynomial (continuous on $[-2,2]$, differentiable on $(-2,2)$), and $f(-2)=0=f(2)$, so Rolle's hypotheses hold.

$f^{\mathrm{\prime }}(x)=2x$, so $f^{\mathrm{\prime }}(c)=0⟹c=0$.

$\frac{d}{dx}\sin (3x)=\cos (3x)\cdot 3=3\cos (3x)$.

${\displaystyle \frac{f(3+h)-f(3)}{h}}={\displaystyle \frac{(3+h{)}^2-9}{h}}={\displaystyle \frac{9+6h+h^2-9}{h}}={\displaystyle \frac{6h+h^2}{h}}=6+h$.

${\displaystyle \underset{h\to 0}{\lim }(6+h)=6}$.

Average rate: ${\displaystyle \frac{f(4)-f(0)}{4-0}}={\displaystyle \frac{16-0}{4}}=4$.

$f^{\mathrm{\prime }}(x)=2x$, so $f^{\mathrm{\prime }}(c)=4⟹2c=4⟹c=2$.

$2\in (0,4)$ ✓.

$f^{\mathrm{\prime }}(x)=5(x^2+1{)}^4\cdot 2x=10x(x^2+1{)}^4$.