Question

how would i differentiate 3^(cos(xsinx))?

Answer 1

Use the chain rule twice.

First, tell me what is the derivative of 3^x ?

Answer 2

That would be ln(3) x 3^x

Answer 3

Yes,

\(\large\color{black}{ \displaystyle \frac{ d }{dx} 3^x=3^x\ln(3) }\)

Answer 4

Then, by the Chain Rule principal you get:

\(\large\color{black}{ \displaystyle \frac{ d }{dx} 3^{f(x)}=3^{f(x)}\ln(3)\times f'(x) }\)

Answer 5

So then:

\(\Large\color{black}{ \displaystyle \frac{ d }{dx} 3^{\cos\left(\sin x\right)}=3^{\cos\left(\sin x\right)}\ln(3)\times ? }\)

Answer 6

You just need to find that chain rule derivative for

\(\Large\color{black}{ \displaystyle {\cos\left(\sin x\right)} }\)

Answer 7

what about with an x in front of the sin, like \[3^{\cos(xsin(x))}\]

Answer 8

Oh, there is an x in front of sin x? Then,

\(\large\color{black}{ \displaystyle \frac{ d }{dx} 3^{\cos\left(x\sin x\right)}=3^{\cos\left(x\sin x\right)}\ln(3)\times \left\{\frac{d}{dx}~\cos\left(x\sin x\right)\right\} }\)

Answer 9

would the derivative of cos(xsinx) be -sin(xsinx)*x*cos(x)?

Answer 10

x*sinx

will go by the product rule

Answer 11

what is the derivative of x•sin(x) ?

Answer 12

1*cos(x), I believe.

Answer 13

\(\large\color{black}{ \displaystyle \frac{ d}{dx}(f\cdot g)=f'g+fg' }\)

Answer 14

\(\large\color{black}{ \displaystyle \frac{ d}{dx}(x\cdot \sin x)=(x')\cdot \sin x+x \cdot (\sin x') }\)

Answer 15

oh, so just sin(x)+(x*cos(x))?

Answer 16

yes,

\(\large\color{black}{ \displaystyle \frac{ d}{dx}(x\cdot \sin x)= \sin x+x \cos x}\)

Answer 17

And therefore:

\(\large\color{black}{ \displaystyle \frac{ d}{dx}~\sin(x\cdot \sin x)=\cos(x\cdot \sin x)\times (\sin x+x \cos x)}\)

Answer 18

If you ever don't get something, please ask...

Answer 19

And lastly, therefore:

\(\normalsize\color{black}{\displaystyle \frac{ d}{dx}~3^{\Large\sin(x\cdot \sin x)}=\ln(3)\cdot 3^{\Large\sin(x\cdot \sin x)}\cdot \color{red}{\cos(x\cdot \sin x)\times (\sin x+x \cos x)}}\)

Answer 20

Thank you so much!

Answer 21

yw