Differentiate the Function: $y=e^{\tan x}$ $y=e^{\tan x}$
The book says to use the Chain Rule. Let $u = \tan x$. Thus, $y = e^u$
$du = \sec^2x\ dx$
$\frac{du}{\sec^2x}= dx$
I am confused at this point. 
The book explains the method of calculating this problem with the following: 
$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}= e^u\frac{du}{dx}=e^{\tan x}\sec^2x \quad (???)$$ 
Maybe I am confusing $u$ substitution with what they are asking me to do. 
What does this line mean in plain English can someone explain? 
 A: In plain english: you're differentiating $e$ to the 'something': your answer is the derivative of 'something' times $e$ to the 'something'.
In this case, your 'something' is $\tan x$. You correctly differentiated that to get $\sec^2 x$, and hence the answer is $e^{\tan x}\sec^2 x$.
EDIT: if you stick in your value of $u$, you get $y=e^u$. At that point, it is clear that $dy/du = e^u$.
You already found that $du = \sec^2 x dx$, so if you divide both sides by $dx$ then you get $du/dx = \sec^2 x$
A: The chain rule says that if $y$ depends on $u$, and $u$ on $x$, then
$$\frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy}{\mathrm du}\cdot \frac{\mathrm du}{\mathrm dx}.$$
In other words,

the rate of change of $y$ with respect to $x$ is the product of two "chained" rates of change (that of $y$ with respect to $u$ and that of $u$ with respect to $x$).

When you calculated the derivative of $u$, you got
$$\mathrm du = sec^2 x \ \mathrm dx,$$
but maybe a clearer way to write that in this context would be
$$\frac{\mathrm du}{\mathrm dx} = sec^2 x.$$
Similarly, since $y=e^u$, we have $\frac{\mathrm dy}{\mathrm du} = e^u.$ Therefore, the result follows, as you have written out at the end of your question.
A: An interesting alternative is to note that $\ln y = \tan x$, so differentiating implicitly with respect to $x$ yields $$\frac{1}{y} \frac{\mathrm{d}y}{\mathrm{d}x} = \sec^2 x \implies \frac{\mathrm{d}y}{\mathrm{d}x} = e^{\tan x} \sec^2 x.$$
NB: This is more of an elegant way to solve the problem, not something I would recommend for you right now, as you learn. 
A: you can go through following steps


*

*Take log with its base as "$e$" on both sides so that on right side you will have only $tan(x)$ term.


*Take derivative on both sides with respect to $x$


*keep derivative term I.e. $dy/dx$ on left side and shift all remaining on right side .your answer is ready

A: First note that
$$\frac{d}{dx}\left[e^x\right]=e^x$$
$$\frac{d}{dx}\left[\tan x\right]=\sec^2 x$$
Also note that the chain rule is
$$ 
\frac{d}{dx}\left[f(g(x))\right]=\frac{d}{dg(x)}\left[f(g(x))\right]\frac{d}{dx}\left[g(x)\right]
$$
Therefore
$$\frac{d}{dx}\left[e^{\tan x}\right]$$
$$=e^{\tan x}\frac{d}{dx}\left[\tan x\right]$$
$$=e^{\tan x}\sec^2 x$$
A: As the given
$$ y = e^{tanx}$$
here put $u =tanx$ then $du=sec^2xdx \implies \frac{du}{dx} = sec^2x$.
Now given eq become
$$y=e^u$$
differentiating w.r.t.x
$$\frac{dy}{dx}=e^u\frac{du}{dx}$$
by putting the values
$$\frac{dy}{dx}=e^{tanx}sec^2x$$
A: 
For computational purpose use this result 

$if y=e^{f(x)}then \frac{dy}{dx}=e^{f(x)}.f^1(x)$
For your question $f(x)=tanx$
$=>f^1(x)=\sec^2x$
So the answer is 
$\frac{dy}{dx}=e^{tanx}(\sec^2x)$
