Differentiation practice questions — A-Level Maths

$f'(x) = \dfrac{(1/x) \cdot x - \ln x \cdot 1}{x^2} = \dfrac{1 - \ln x}{x^2}$. At $x = 1$: $\dfrac{1 - 0}{1} = 1$.

Outer: $\sec^2(x^2)$. Inner derivative: $2x$. So $f'(x) = 2x\sec^2(x^2)$.

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

$f'(x) = 2x$, so $f'(c) = 0 \implies c = 0$.

$f'(x) = 5(x^2 + 1)^4 \cdot 2x = 10x(x^2 + 1)^4$.

$f'(x) = 2x \ln x + x^2 \cdot \tfrac{1}{x} = 2x \ln x + x = x(2\ln x + 1)$.

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

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

So $$f'(x) = 8(2x-1)^3(x+1) + (2x-1)^4.$$

Factor out $(2x-1)^3$: $$f'(x) = (2x-1)^3\big[\,8(x+1) + (2x-1)\,\big] = (2x-1)^3(10x+7).$$

Substitute $x=1$, where $2x-1=1$: $$f'(1) = (1)^3(10+7) = 17.$$

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

With $u = 2x - 1$ and $v = x^2 + 1$, we have $u' = 2$ and $v' = 2x$, so $$f'(x) = \frac{2(x^2 + 1) - (2x - 1)(2x)}{(x^2 + 1)^2}$$ 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 $$f'(1) = \frac{2}{4} = \frac{1}{2}$$

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

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

$f'(x) = (x)' e^x + x (e^x)' = e^x + x e^x = e^x(x + 1)$.