frac{d}{d x}left(frac{2^x+3^x}{4^x}right) = | vedclass

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Question

(B) Let $y = \frac{2^x + 3^x}{4^x} = \frac{2^x}{4^x} + \frac{3^x}{4^x} = \left(\frac{2}{4}\right)^x + \left(\frac{3}{4}\right)^x = \left(\frac{1}{2}\r

Vedclass · Accepted Answer

$\left(\frac{1}{2}\right)^x \log \frac{1}{2} + \left(\frac{3}{4}\right)^x \log \frac{3}{4}$

Vedclass · Answer

(B) Let $y = \frac{2^x + 3^x}{4^x} = \frac{2^x}{4^x} + \frac{3^x}{4^x} = \left(\frac{2}{4}\right)^x + \left(\frac{3}{4}\right)^x = \left(\frac{1}{2}\right)^x + \left(\frac{3}{4}\right)^x$. Now,differentiate with respect to $x$ using the formula $\frac{d}{dx}(a^x) = a^x \log a$: $\frac{dy}{dx} = \frac{d}{dx}\left(\left(\frac{1}{2}\right)^x\right) + \frac{d}{dx}\left(\left(\frac{3}{4}\right)^x\right)$ $\frac{dy}{dx} = \left(\frac{1}{2}\right)^x \log \frac{1}{2} + \left(\frac{3}{4}\right)^x \log \frac{3}{4}$. Thus,the correct option is $B$.

$\frac{d}{d x}\left(\frac{2^x+3^x}{4^x}\right) = $ . . . . . .

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