Question

If $y=x^{-\frac{1}{2}}+{\log }_{5}x+\frac{\sin x}{\cos x}+2^x$, then find $\frac{dy}{dx}$

• A. $-\frac{1}{2}{x}^{-3/2}+\frac{1}{x{\log }_{e}5}+{\sec }^{2}x+2^x\log 2$
• B. $\frac{1}{2}{x}^{-3/2}+\frac{1}{x{\log }_{e}5}+{\sec }^{2}x+2^x\log 2$
• C. $-\frac{3}{2}{x}^{-3/2}+\frac{1}{x{\log }_{e}5}+{\sec }^{2}x+2^x\log 2$
• D. $-\frac{1}{2}{x}^{-3/2}+\frac{1}{x{\log }_{e}5}+{\cos }^{2}x+2^x\log 2$

Answer 1

The correct option is A $-\frac{1}{2}{x}^{-3/2}+\frac{1}{x{\log }_{e}5}+{\sec }^{2}x+2^x\log 2$

Here, we have function $y=x^{-1/2}+{\log }_{5}x+\tan x+2^x$

On differentiating w.r.t x, we get

$\frac{dy}{dx}=\frac{d}{dx}(x{)}^{-1/2}+\frac{d}{dx}({\log }_{5}x)+\frac{d}{dx}\tan x+\frac{d}{dx}(2^x)$

$=-\frac{1}{2}(x{)}^{-1/2-1}+\frac{1}{x{\log }_{e}5}+{\sec }^{2}x+2^x\log 2$

$=-\frac{1}{2}{x}^{-3/2}+\frac{1}{x{\log }_{e}5}+{\sec }^{2}x+2^x\log 2$