Q.Find the second order derivative of the function: $x^{20}$

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

Factor the common $e^{-3x}$:

Comparing with $e^{\alpha x}(4e^{2x}-8e^{x}-9)$ gives $\alpha=-3$.

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

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

Since $\dfrac{d^2y}{dx^2}=-x^{-3/2}=-8\left(\dfrac{1}{8}x^{-3/2}\right)$, we get

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 $-2xy$).

Step 3 — form the required combination.

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

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

$g'(x)=80x^3+70\sin x+80\cos 2x$.

$g''(x)=240x^2+70\cos x-160\sin 2x$.

$g'''(x)=480x-70\sin x-320\cos 2x$.

$g'''(0)=0-0-320(1)=-320$.

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

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

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

Multiply by $\dfrac{dx}{dy}=\dfrac{1}{\cos x+e^x}$: $\dfrac{d^2x}{dy^2}=\dfrac{\sin x-e^x}{(\cos x+e^x)^3}$.

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

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

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

At $x=1$: $f'(1)=27e^5$ and $f''(1)=145e^5$.

So $f''(1)-10f'(1)=145e^5-270e^5=-125e^5$.

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

Then

and

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

Differentiating again,

At $x=3$: $\dfrac{1}{3-1}=\dfrac{1}{2}.$

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. $c=6$.

(For consistency: $f'(x)=3x^2+2ax+b\Rightarrow f'(1)=3+2a+b=a$; $f''(x)=6x+2a\Rightarrow f''(2)=12+2a=b$. Solving gives $a=-5,\;b=2$, and $c=6$.)

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: $\frac{d}{dx}(12x^{-6})=12(-6)x^{-7}=-72x^{-7}$.

Common Mistakes

• Sign slips when differentiating negative powers; keep track that each step lowers the exponent by one.

ANSWER: E