Question

13.6 Determine the following derivatives with help of the chain rule, maybe applied several times.
In each case, write down the inner and outer functions used.
a) \(\Large h(x) = cos(sin x) \)

Answer 1

?? what inner and outer functions

Answer 2

i am not sure what it is but i asked before here this kind of question there were people who was familiar with it

Answer 3

f(x) = cos(x)
g(x) = sin(x)
h(x) = f(g(x)) perhaps

Answer 4

ok to start with it good for me maybe its right i ask some more people to be sure.. thanks

Answer 5

np :)

Answer 6

\[-\sin (\sin(x))*\cos(x)\]
Where -sin(sin(x) is the outer function in respect to the inner function and cos(x) is the inner function

Answer 7

@mahmit2012 what you think about Spacelimbus's solution?

Answer 8

thank you Spacelimbus :) i dont know too what the solution is, lets wait.. what it comes out

Answer 9

h(x) =cos(sinx) , i think the inner f is sinx and the outer f is cos.
now by chain rule, let u=sinx , then du/dx = cosx
therefore cos(sinx) becomes cosu, and dy/du =-sinu
then dy/dx =dy/du *du/dx,just substute the values. nb h(x) = y

Answer 10

ok thank you pota
by the way your son - daughter nice sweet baby

Answer 11

Here you have the solution again computed by WolframAlpha. With Leibniz notation it is easier to see what you are doing in this case.

\[h(x)=\cos(\sin(x))\]

now let u=sin(x) then \[du/dx = \cos(x)\]
and

\[dy/du= -\sin(x)\]
so that \[du/dx*dy/du= dy/dx = -\sin(u)\cos(x) = -\sin(\sin(x)\cos(x)\]

Answer 12

http://www.wolframalpha.com/input/?i=cos%28sinx%29+derive

Answer 13

thank you very much Spacelimbus it looks good

Answer 14

you are welcome