Q.If $y = \sin(\sin x)$, prove that $\dfrac{d^2 y}{dx^2} + \tan x\, \dfrac{dy}{dx} + y\cos^2 x = 0$.

With $y=4e^{-x}-2e^{-2x}-e^{-3x}$:

Factor the common $e^{-3x}$: …

With $y=4x^{1/2}$:

Also $y^2=16x\Rightarrow \dfrac{1}{y^2}=\dfrac{1}{16x}$. Then

$\frac{1}{y^2}\frac{dy}{dx}=\frac{2x^{-1/2}}{16x}=\frac{1}{8}x^{-3/2}.$ …

Concept. Because $y=e^{-x^2}$ satisfies $y'=-2xy$, its higher derivatives can be written back in terms of $y$ itself — a differential-equation viewpoint that makes the given combination easy to evaluate.

Step 1 — first derivative (chain rule).

Step 2 — second derivative (product rule on $-2xy$). …

Cofactors of row 1: $\det\begin{pmatrix}4&3\\4&8\end{pmatrix}=20$, $-\det\begin{pmatrix}-8&3\\2&8\end{pmatrix}=-(-70)=70$, $\det\begin{pmatrix}-8&4\\2&4\end{pmatrix}=-40$.

$g(x)=x^4(20)+(-\cos x)(70)+(-\sin 2x)(-40)=20x^4-70\cos x+40\sin 2x$. …

$\dfrac{dy}{dx}=\cos x+e^x\Rightarrow \dfrac{dx}{dy}=\dfrac{1}{\cos x+e^x}$.

$\dfrac{d^2x}{dy^2}=\dfrac{d}{dy}\!\left(\dfrac{dx}{dy}\right)=\dfrac{d}{dx}\!\left(\dfrac{1}{\cos x+e^x}\right)\cdot\dfrac{dx}{dy}$. …

Given $f(x)=(2x+3)e^{5x}$.

First derivative: $f'(x)=2e^{5x}+5(2x+3)e^{5x}=e^{5x}(10x+17)$.

Second derivative: $f''(x)=5e^{5x}(10x+17)+10e^{5x}=e^{5x}(50x+95)$. …

With $f(x)=x^{-2}$:

Then

and …

With $y=(x-1)\log_e(x-1)$,

Differentiating again, …

Write $a=f'(1),\;b=f''(2),\;c=f'''(3)$ (all constants), so $f(x)=x^3+ax^2+bx+c$.

Then $f'''(x)=6$ for all $x$, hence $f'''(3)=6$, i.e. $c=6$. …

Given $y=(5x-2)e^x$. First derivative:

Second derivative: …

Concept and Intuition

Apply the power rule $\frac{d}{dx}x^n=nx^{n-1}$ repeatedly, working from the inside out.

Step-by-Step Solution

1. $\frac{d}{dx}x^{-3}=-3x^{-4}$, then $\frac{d^2}{dx^2}x^{-3}=\frac{d}{dx}(-3x^{-4})=12x^{-5}$.
2. Multiply by $\frac1x$: $\frac1x\cdot12x^{-5}=12x^{-6}$.
3. Differentiate: $\frac{d}{dx}(12x^{-6})=12(-6)x^{-7}=-72x^{-7}$. …