Question

how to find the derivative of y=e^cos(sqrt3x)?

Answer 1

We have:
\[{y=e^{\cos(\sqrt{3x})}}\] Right?

Answer 2

yes that is correct

Answer 3

Then we can proceed:
\[\frac{d}{dx}e^{\cos(\sqrt{3x})}=\frac{d}{dx}e^{\cos\left(\sqrt{f(x)}\right)}=\frac{d}{dx}e^{\cos[g(f(x))]}=\frac{d}{dx}e^{h[g(f(x))]}=\frac{d}{dx}j\bigg({h\Big[g[f(x)]\Big]}\bigg)\]
Using the chain rule, we can simplify this

Answer 4

Make Sense?

Answer 5

yes

Answer 6

OKAY! SO lets start. Through that little simplification above, we obtained two pieces of information: One is that:
\[\frac{d}{dx}e^{\cos(\sqrt{3x})}=\frac{d}{dx}j\bigg({h\Big[g[f(x)]\Big]}\bigg)\]
And the second is a result of the first:
\[\eqalign{
&f(x)=3x \\
&g(x)=\sqrt{x} \\
&h(x)=cos(x) \\
&j(x)=e^x \\
}\]
So then we can use the chain rule to figure out:
\[\frac{d}{dx}j\bigg({h\Big[g[f(x)]\Big]}\bigg)=j\phantom{.}'\bigg({h\Big[g[f(x)]\Big]}\bigg)\times h'\Big[g[f(x)]\Big]\times g'[f(x)]\times f'(x)\]
So let us find the derivatives for each!
\[\eqalign{
&f(x)=3x\phantom{spc}\rightarrow f'(x)=3\\
&g(x)=\sqrt{x}\phantom{spc}\rightarrow g'(x)=\frac{1}{2\sqrt{x}}\\
&h(x)=cos(x)\phantom{spc}\rightarrow h'(x)=-sin(x)\\
&j(x)=e^x\phantom{spc}\rightarrow j'(x)=e^x\\
}\]
Now it's just a little algebra. Kinda straightforward, but kinda messy too haha

Answer 7

\[\eqalign{
&j\phantom{.}'\bigg({h\Big[g[f(x)]\Big]}\bigg)\times h'\Big[g[f(x)]\Big]\times g'[f(x)]\times f'(x) \\
=&e^{\cos(\sqrt{3x})}\times -\sin(\sqrt{3x})\times\frac{1}{2\sqrt{3x}}\times3 \\
=&\frac{-3e^{\cos(\sqrt{3x})}\sin\sqrt{3x}}{2\sqrt{3x}}\\
}\]

Answer 8

is that the answer??

Answer 9

Sorry but his solution is too complicated. Look at the derivative of the exponential function. \[(e^{x} )\prime = e^{x} \times x \prime = e^{x} \times 1 =e^{x}\]

Answer 10

Yes that's the final answer

Answer 11

@KeithAfasCalcLover your solution is complex but your answer is right.

Answer 12

The function is complex in the sense that it is the composite of four functions and so If the function is complex, so will the answer to give a THOUGHROUGH answer.

Answer 13

Wow you seem to have a deep understanding of this. I didn't think of it from that point of view.

Answer 14

Is that irony? Besides, my answer wasn't even that complex. Just emphasizing the chain rule with more than two chain links. And I thought it was pretty thorough...

Answer 15

Usually when I take derivatives of exponential functions I just use the basic exponential function as a reference.

Answer 16

And what if they have a inner function?

Answer 17

Resort to decomposing. lol