Differentiate the following w.r.t $x$ :$\large \frac{8^x}{x^8}$

Question

$\begin{array}{1 1} (A)\;\frac{8^x}{x^8}[\log 8+\frac{8}{x}] \\ (B)\;\frac{8^x}{x^8}[x\log 8+\frac{8}{x}] \\ (C)\;\frac{8^x}{x^8}[\log 8-\frac{8}{x}] \\ (D)\;\frac{8^x}{x^8}[x\log 8-\frac{8}{x}] \end{array} $

Answer 1

Toolbox:

• Product rule : $\large\frac{d}{dx}$$(uv)=u\large\frac{d}{dx}$$(v)+v\large\frac{d}{dx}$$(u)$.$(i.e)$ $(uv)'=uv'+vu'$
• $\large\frac{d}{dx}$$(a^x)=a^x\log a$
• $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$

$y=\large\frac{8^x}{x^8}$

$\;\;=8^xx^{-8}$

This can be differentiated by using the product rule .

$d(uv)=uv'+vu'$

Here $u=8^x,v=x^{-8}$

$u'=8^x\log 8,v'=-8x^{-9}$

$\large\frac{dy}{dx}$$=8^x.(-8x^{-9})+x^{-8}.\log 8$

$\quad=\large\frac{-8.8^x}{x^9}+\frac{8^x\log 8}{x^8}$

$\quad=\large\frac{8^{\large x}}{x^{\large 8}}$$[\log 8-\large\frac{8}{x}]$