Differentiation practice questions — A-Level Maths

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

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

$f$ 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 $f^{\mathrm{\prime }}(c)=0⟹c=0$.

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

$f^{\mathrm{\prime }}(x)=2x\ln x+x^2\cdot \frac{1}{x}=2x\ln x+x=x(2\ln x+1)$.

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

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

With $u=2x-1$ and $v=x^2+1$, we have $u^{\mathrm{\prime }}=2$ and $v^{\mathrm{\prime }}=2x$, so $$f^{\mathrm{\prime }}(x)=\frac{2(x^2+1)-(2x-1)(2x)}{(x^2+1{)}^2}$$ At $x=1$ the numerator is $$2(2)-(1)(2)=4-2=2$$ and the denominator is $$(1^2+1{)}^2=2^2=4$$ Therefore $$f^{\mathrm{\prime }}(1)=\frac{2}{4}=\frac{1}{2}$$

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

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

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