In Calculus differentiate f(x)=(e^sinx) + (sin(e^x))?

Question

I am in calculus 1 and am having a little bit of a hard time. we just learned about the chain rule.

Answer 1

here's how you do the chain rule:

it is defined as df/du*du/dx. Basically, it is used for functions with complex terms where you can't take the derivative directly. You need to take the derivative in multiple steps.

Here, let's differentitate e^[sinx]. Let's call the function 'f'. Now let's define u = sinx. Thus the function simplifies to e^u. Taking the derivative of this function with respect to 'u' is simple, it is e^u. Now let's take the derivative of sinx with respect to 'x'. It is cosx.

Using df/du*du/dx, we get the derivative as e^(sinx)*cosx

Using the same procedure, you can find the derivative of sin(e^x). We defined u = (e^x)

Thus, df/du = df(sin'u') = cos 'u'; du/dx = du(e^x) = e^x

Simplifying, the derivative is (e^x)*cos(e^x)

Adding the two derivative gives the derivative of the overall function:

f'(x) = e^(sinx)*cosx + (e^x)*cos(e^x)

Hope this helps

Answer 2

Yeah, a typical chain rule problem!

You can write both terms as the "chain" f(g(x)), with derivative
f'(g(x)) * g'(x).
(Maybe you learn y = f(u) and u = g(x), and then the derivative is f'(u) * g'(x). The principle is the same.)

For the first term, take f(u) = e^u and g(x) = sin x. The derivative:
f'(u) * g(x) = e^u * cos x = e^(sin x) * cos x

For the second term, f(u) = sin u and g(x) = e^x. The derivative:
f'(u) * g(x) = cos u * e^x = cos (e^x) * e^x

So the derivative of your function is
f'(x) = e^(sin x) * cos x + cos (e^x) * e^x

which really cannot be simplified much further.

Answer 3

use the chain rule. take the by-manufactured from the exterior function mutually as leaving the interior on my own and then take the by-manufactured from the interior function. by-manufactured from ln(x) = a million/x, so by-manufactured from ln(3x-4) is a million/(3x-4) * 3 = 3 / (3x-4)

Answer 4

http://www.mathworld.wolfram.com
http://archives.math.utk.edu/visual.calculus/2/chain_rule.4/index.html
http://www.calculus-help.com/funstuff/tutorials/derivatives/deriv05.html

A great math resource!

Answer 5

cos(x)(e^sinx) + e^xcos(e^x)

Answer 6

f'(x)
=e^(sinx)*cosx+cos(e^x)*e^x