Differentiation rules and standard derivatives: practice worksheet

1. Question 1warm-up · no calculator

   No calculator. Let $f(x)=3x^{4/3}-6x^{-1/2}+5$, where $x>0$.

   (a)Find $f'(x)$.[2]

   Markscheme

   Attempts to apply the power rule to both non-constant terms.	M1
   $f'(x)=4x^{1/3}+3x^{-3/2}$.	A1

   Accept equivalent forms.

   Answer: $f'(x)=4x^{1/3}+3x^{-3/2}$.

   (b)Find $f''(1)$.[1]

   Markscheme

   $f''(1)=\frac{4}{3}-\frac{9}{2}=-\frac{19}{6}$.

   A1

   FT their derivative from part (a), provided it contains a non-constant term and is correctly differentiated and evaluated.

   Answer: $f''(1)=-\frac{19}{6}$.

2. Question 2warm-up · no calculator

   No calculator. A curve has equation $y=e^x\sin(2x)+\cos x$.

   (a)Find $\frac{dy}{dx}$.[3]

   Markscheme

   Attempts the product rule on $e^x\sin(2x)$.	M1
   $\frac{d}{dx}\bigl(e^x\sin(2x)\bigr)=e^x\sin(2x)+2e^x\cos(2x)$.	A1
   $\frac{dy}{dx}=e^x\sin(2x)+2e^x\cos(2x)-\sin x$.	A1

   Accept a correctly factorised equivalent expression.

   Answer: $\frac{dy}{dx}=e^x\sin(2x)+2e^x\cos(2x)-\sin x$.

   (b)Hence find the gradient of the curve at $x=0$.[1]

   Markscheme

   The gradient is $1(0)+2(1)(1)-0=2$.

   A1

   FT correct evaluation of their derivative from part (a).

   Answer: $2$.

3. Question 3exam-style · no calculator

   No calculator. Let $f(x)=\ln\left(\frac{x+2}{2x+1}\right)$, where $x>0$.

   (a)Show that $f'(x)=-\frac{3}{(x+2)(2x+1)}$.[4]

   Markscheme

   Writes $f(x)=\ln(x+2)-\ln(2x+1)$ and attempts to differentiate.	M1
   $f'(x)=\frac{1}{x+2}-\frac{2}{2x+1}$.	A1
   $f'(x)=\frac{(2x+1)-2(x+2)}{(x+2)(2x+1)}$.	A1
   Simplifies the numerator: $2x+1-2x-4=-3$.	A1
   $f'(x)=-\frac{3}{(x+2)(2x+1)}$.	AG

   Alternative method: award M1 for attempting the chain and quotient rules, A1 for the inner derivative $-3/(2x+1)^2$, A1 for multiplying by $(2x+1)/(x+2)$, and A1 for cancelling factors. The simplification must be shown.

   Answer: $f'(x)=-\frac{3}{(x+2)(2x+1)}$.

   (b)Hence find the gradient of the graph of $f$ at $x=1$.[1]

   Markscheme

   $f'(1)=-\frac{3}{3\times3}=-\frac{1}{3}$.

   A1

   Answer: $-\frac{1}{3}$.

   (c)Find $f''(1)$.[2]

   Markscheme

   Differentiates $-3(2x^2+5x+2)^{-1}$ to obtain $f''(x)=\frac{3(4x+5)}{(2x^2+5x+2)^2}$.	M1
   $f''(1)=\frac{27}{81}=\frac{1}{3}$.	A1

   For M1, accept a correct quotient-rule attempt with an error in simplification. Accept equivalent forms.

   Answer: $f''(1)=\frac{1}{3}$.

4. Question 4exam-style · no calculator

   No calculator. Let $u(x)=2^{3x}$ and $v(x)=\log_2 x$, where $x>0$. Define $g(x)=u(x)v(x)$.

   (a)Write down $u'(x)$ and $v'(x)$.[2]

   Markscheme

   $u'(x)=3(\ln 2)2^{3x}$.	A1
   $v'(x)=\frac{1}{x\ln 2}$.	A1

   Answer: $u'(x)=3(\ln 2)2^{3x}$ and $v'(x)=\frac{1}{x\ln 2}$.

   (b)Show that $g'(x)=2^{3x}\left(3\ln x+\frac{1}{x\ln 2}\right)$.[3]

   Markscheme

   Uses $g'(x)=u'(x)v(x)+u(x)v'(x)$.	M1
   $g'(x)=3(\ln 2)2^{3x}\frac{\ln x}{\ln 2}+\frac{2^{3x}}{x\ln 2}$.	A1
   Cancels $\ln 2$ in the first term and factorises to obtain $2^{3x}\left(3\ln x+\frac{1}{x\ln 2}\right)$.	A1
   $g'(x)=2^{3x}\left(3\ln x+\frac{1}{x\ln 2}\right)$.	AG

   The cancellation and factorisation must be evident.

   Answer: $g'(x)=2^{3x}\left(3\ln x+\frac{1}{x\ln 2}\right)$.

   (c)Find $g''(1)$, giving your answer in exact form.[3]

   Markscheme

   Applies the product rule to the expression for $g'(x)$.	M1
   $g''(x)=2^{3x}\left(9(\ln 2)\ln x+\frac{6}{x}-\frac{1}{x^2\ln 2}\right)$.	A1
   $g''(1)=8\left(6-\frac{1}{\ln 2}\right)=48-\frac{8}{\ln 2}$.	A1

   Accept an unsimplified correct expression for $g''(x)$. FT the final evaluation mark from their second derivative, provided the product rule has been attempted.

   Answer: $g''(1)=48-\frac{8}{\ln 2}$.

5. Question 5exam-style · no calculator

   No calculator. Let $h(x)=\frac{\sec x}{1+\arctan(\sin x)}$, where $-\frac{\pi}{2}<x<\frac{\pi}{2}$. Define $d(x)=1+\arctan(\sin x)$, so that $d(x)h(x)=\sec x$.

   (a)Show that $h'(x)=\frac{\sec x\left[(1+\arctan(\sin x))\tan x-\frac{\cos x}{1+\sin^2 x}\right]}{(1+\arctan(\sin x))^2}$.[5]

   Markscheme

   Attempts the quotient rule with numerator $\sec x$ and denominator $d(x)$.	M1
   $\frac{d}{dx}(\sec x)=\sec x\tan x$.	A1
   $d'(x)=\frac{\cos x}{1+\sin^2 x}$.	A1
   $h'(x)=\frac{d(x)\sec x\tan x-\sec x\frac{\cos x}{1+\sin^2 x}}{d(x)^2}$.	A1
   Factorises $\sec x$ from the numerator and substitutes $d(x)=1+\arctan(\sin x)$.	A1
   $h'(x)=\frac{\sec x\left[(1+\arctan(\sin x))\tan x-\frac{\cos x}{1+\sin^2 x}\right]}{(1+\arctan(\sin x))^2}$.	AG

   Accept a product-rule approach using $h(x)=\sec x\,[d(x)]^{-1}$ with corresponding accuracy steps. The factorisation must be shown.

   Answer: $h'(x)=\frac{\sec x\left[(1+\arctan(\sin x))\tan x-\frac{\cos x}{1+\sin^2 x}\right]}{(1+\arctan(\sin x))^2}$.

   (b)Hence find $h'(0)$.[1]

   Markscheme

   $h'(0)=\frac{1(1\times0-1)}{1^2}=-1$.

   A1

   Answer: $h'(0)=-1$.

   (c)By differentiating the identity $d(x)h(x)=\sec x$ twice, find $h''(0)$.[3]

   Markscheme

   Applies the product rule twice to obtain $dh''+2d'h'+d''h=\sec x\tan^2 x+\sec^3 x$.	M1
   Uses $d(0)=1$, $d'(0)=1$ and $d''(0)=0$ to obtain $h''(0)+2(-1)=1$.	A1
   $h''(0)=3$.	A1

   The value $d''(0)=0$ follows by differentiating $d'(x)=\cos x/(1+\sin^2 x)$. FT their value from part (b) in the last two lines, giving $h''(0)=1-2h'(0)$.

   Answer: $h''(0)=3$.

6. Question 6challenge · no calculator

   No calculator. This question links differentiation to equations of perpendicular lines. For $x>0$ and $k\in\mathbb{R}$, define $u(x)=\frac{x-1}{x+1}$ and $F_k(x)=\arcsin(u(x))+k\arccos(u(x))$. The tangent to the graph of $F_k$ at $x=1$ is denoted by $T_k$.

   (a)Show that $1-[u(x)]^2=\frac{4x}{(x+1)^2}$.[2]

   Markscheme

   Writes $1-[u(x)]^2=\frac{(x+1)^2-(x-1)^2}{(x+1)^2}$.	M1
   Expands the numerator to obtain $x^2+2x+1-(x^2-2x+1)=4x$.	A1
   $1-[u(x)]^2=\frac{4x}{(x+1)^2}$.	AG

   Accept factorisation of the difference of two squares instead of expansion.

   Answer: $1-[u(x)]^2=\frac{4x}{(x+1)^2}$.

   (b)Hence show that $F_k'(x)=\frac{1-k}{(x+1)\sqrt{x}}$.[4]

   Markscheme

   Attempts the chain rule for both inverse trigonometric terms.	M1
   $u'(x)=\frac{(x+1)-(x-1)}{(x+1)^2}=\frac{2}{(x+1)^2}$.	A1
   $F_k'(x)=\frac{u'(x)}{\sqrt{1-u(x)^2}}-\frac{ku'(x)}{\sqrt{1-u(x)^2}}=\frac{2(1-k)}{(x+1)^2\sqrt{1-u(x)^2}}$.	A1
   Since $x>0$, uses $\sqrt{1-u(x)^2}=\frac{2\sqrt{x}}{x+1}$ and cancels factors to obtain $\frac{1-k}{(x+1)\sqrt{x}}$.	A1
   $F_k'(x)=\frac{1-k}{(x+1)\sqrt{x}}$.	AG

   The negative sign in the derivative of arccos must be evident. Accept first using $\arcsin u+\arccos u=\pi/2$, with corresponding method and accuracy steps.

   Answer: $F_k'(x)=\frac{1-k}{(x+1)\sqrt{x}}$.

   (c)Find an equation of $T_k$ in terms of $k$.[3]

   Markscheme

   Uses the tangent equation $y-F_k(1)=F_k'(1)(x-1)$.	M1
   Obtains $F_k(1)=\frac{k\pi}{2}$ and $F_k'(1)=\frac{1-k}{2}$.	A1
   $y-\frac{k\pi}{2}=\frac{1-k}{2}(x-1)$.	A1

   Accept equivalent forms of the line equation.

   Answer: $y-\frac{k\pi}{2}=\frac{1-k}{2}(x-1)$.

   (d)Given that $T_k$ is perpendicular to the line $y=2x-3$, determine $k$ and the $x$-intercept of $T_k$.[3]

   Markscheme

   Uses perpendicular gradients to set $\frac{1-k}{2}=-\frac{1}{2}$.	M1
   $k=2$.	A1
   Sets $y=0$ in $y-\pi=-\frac{1}{2}(x-1)$ to obtain $x=1+2\pi$.	A1

   FT the final mark from their value of $k$ and their tangent equation, provided the resulting line has a unique $x$-intercept.

   Answer: $k=2$; the $x$-intercept is $1+2\pi$, corresponding to the point $(1+2\pi,0)$.

Original questions and notes written by Pietro Meloni, PhD, from the IB syllabus. Not endorsed by or affiliated with the International Baccalaureate Organization. Found a mistake? Please tell me privately at pietro@pietromeloni.com. More questions on this topic: Calculus practice.