A Clean Answer That Is Wrong
A student differentiates $y=x^2\ln x$ by multiplying the two derivatives. Diagnose each line below: click Correct step or Wrong step.
Reality check: try $f=g=x$. $(x\cdot x)'=2x$, but $f'g'=1\cdot 1=1$. Multiplying derivatives is invalid — a product needs the product rule.
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. $\dfrac{d}{dx}\big(x^2\big)$ = ?
R2. $\dfrac{d}{dx}\big(\ln x\big)$ = ?
R3. Is $(fg)' = f'g'$ always true?
The counterexample $f=g=x$ shows the naive guess fails. The correct rule cross-multiplies: each function's rate carries the whole of the other.
Say it as: the first times the derivative of the second, plus the second times the derivative of the first. Both original functions stay intact.
| Function $y$ | Derivative $\dfrac{dy}{dx}$ |
|---|---|
| $x^2\ln x$ | $2x\ln x+x$ |
| $x e^{4x}$ | $e^{4x}(1+4x)$ |
| $x\sin x$ | $\sin x+x\cos x$ |
| $x\tan x$ | $\tan x+x\sec^{2}x$ |
| $e^x\sin x$ | $e^x(\sin x+\cos x)$ |
When one factor is itself a composite, differentiate it with the chain rule first.
$u=x^2,\ u'=2x$; $v=(2x+1)^4,\ v'=8(2x+1)^3$ (chain rule on the factor).
$\dfrac{dy}{dx}=2x(2x+1)^4+x^2\cdot 8(2x+1)^3=2x(2x+1)^3(6x+1)$
The last step pulls out the common factor $2x(2x+1)^3$ and tidies the bracket.
Groups of 4: tap a product on the left, then its derivative on the right — a coloured line links each pair. Six products covering e, ln, trig and polynomial factors. Wrong pairs turn red to fix.
Each attempt below claims a derivative. Decide whether the product rule was applied correctly, then explain your verdict out loud to a partner.
Differentiate $y=x^2\ln(4x)$.
$u=x^2$ ($u'=2x$) and $v=\ln(4x)$ ($v'=\dfrac{4}{4x}=\dfrac{1}{x}$).
$\dfrac{dy}{dx}=2x\ln(4x)+x^2\cdot\dfrac{1}{x}=2x\ln(4x)+x$.
Differentiate $y=x e^{4x}$.
$u=x$ ($u'=1$) and $v=e^{4x}$ ($v'=4e^{4x}$).
$\dfrac{dy}{dx}=1\cdot e^{4x}+x\cdot 4e^{4x}=e^{4x}(1+4x)$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. $\dfrac{d}{dx}\big(x^2\ln x\big)$ = ?
Q2. $\dfrac{d}{dx}\big(x e^{2x}\big)$ = ?
Q3. $\dfrac{d}{dx}\big(x^2\sin x\big)$ = ?
Q4. $\dfrac{d}{dx}\big((x+1)(x-2)\big)$ = ?
Work each on paper, then reveal the solution to check your working.
Differentiate $y=x^5(3x^2+4x-7)$.
Reveal Solution
$\dfrac{dy}{dx}=5x^4(3x^2+4x-7)+x^5(6x+4)=21x^6+24x^5-35x^4=x^4(21x^2+24x-35)$.
Differentiate $y=x^2(2x+1)^4$.
Reveal Solution
$\dfrac{dy}{dx}=2x(2x+1)^4+x^2\cdot 8(2x+1)^3=2x(2x+1)^3(6x+1)$.
For $y=x\cos x$, find the gradient of the curve at $x=\pi$.
Reveal Solution
$\dfrac{dy}{dx}=\cos x-x\sin x$, so at $x=\pi$: $\cos\pi-\pi\sin\pi=-1$.
Given $f(x)=x\ln x$, find the exact coordinates of its stationary point.
Reveal Solution
$f'(x)=\ln x+x\cdot\dfrac{1}{x}=\ln x+1$. Set $\ln x+1=0\ \Rightarrow\ x=e^{-1}=\dfrac{1}{e}$, and $f\!\left(\tfrac{1}{e}\right)=\dfrac{1}{e}\ln\!\left(\tfrac{1}{e}\right)=-\dfrac{1}{e}$. Stationary point $\left(\dfrac{1}{e},-\dfrac{1}{e}\right)$.
Exit Ticket
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