How to find derivative of the following function: $y = 2^{\sqrt{\tan x}}$ , $y' = ?$ I did the following $$\frac{d}{dx}2^{\sqrt{\tan x}} = 2^{\sqrt{\tan x}}\ln{2}(\sqrt{\tan}x)'$$ and stopped here. Can you guide me?
The solution is as follows in the book: $$2^{\sqrt{\tan x}}\frac{\ln{2}}{2\cos^{2}{x}\sqrt{\tan{x}}}$$
$y=e^{\ln 2\sqrt{\tan x}}$. Differentiating using the chain rule, we get $y'= \ln 2 \times {\sqrt{tan x}}'e^{\ln 2\sqrt{\tan x}}=\ln(2)\frac{\sec^{2}x}{2\sqrt{\tan x}}e^{\ln 2\sqrt{\tan x}}$
HINT:
$$\frac{d(a^x)}{dx}=a^x\log a$$
Now apply the chain rule and you should gave it!
HINT 2:
$$\frac{d\sqrt{\tan x}}{dx}=\frac12 (\tan x)^{-1/2}\sec^2 x$$
If $y = 2^{\sqrt{\tan x}}$ then $\displaystyle y = e^{\sqrt{\tan x} \ln 2}$. So, using the chain rule yields $$y' = \frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{\tan x} \cdot \ln 2\right) \cdot e^{\sqrt{\tan x} \ln 2}$$
But, using the chain rule again yields $$\frac{\mathrm{d}}{\mathrm{d}x} \left( \sqrt{\tan x} \cdot \ln 2\right) = \frac{\sec^2 x \ln 2}{2\sqrt{\tan x}}$$
So $$y' = e^{\sqrt{\tan \left(x\right)}\cdot \ln 2}\cdot \frac{\sec ^2\left(x\right)\ln \left(2\right)}{2\sqrt{\tan x}}$$
Changing back to base 2 and simplifying gives $$y'=2^{\sqrt{\tan x}}\frac{\ln{2}}{2\sqrt{\tan{x}}} \cdot \frac{1}{\cos^2 x} \\ =2^{\sqrt{\tan x}}\dfrac{\ln 2}{2\cos^2 x\sqrt{\tan x}}$$
Edit: For added clarification, let's look at $\dfrac{\mathrm{d}}{\mathrm{d}x} \sqrt{\tan x}$. Let $g(x) = \tan x$ and $f(x) = \sqrt{x}$. So $\sqrt{\tan x}$ is the composite function $f(g(x))$. It's derivative is hence $$f'(g(x)) \cdot g'(x) = \frac{1}{2\sqrt{\tan x}} \cdot \sec^2 x$$
$$y={ 2 }^{ \sqrt { tanx } }\quad \Rightarrow \quad lny=\sqrt { tanx } ln2\quad \Rightarrow \quad lny=\left( \sqrt { tanx } ln2 \right) \Rightarrow \quad derivate\quad both\quad sites\quad \frac { y\prime }{ y } =\frac { ln2 }{ 2\sqrt { tanx } \cos { ^{ 2 }x } } \Rightarrow \\ y\prime =\frac { { 2 }^{ \sqrt { tanx } -1 } }{ \sqrt { tanx } \cos { ^{ 2 }x } } ln2$$
