# y=f(sinx)

• Jan 14th 2013, 06:16 AM
Boo
y=f(sinx)
Hello!

then
y'(x) should be f'(sinx)cosx
y''(x)= f''(sinx) cos x+f'(sinx)(-sinx)
y'''(x)= f'''(x)-2f''(sinx)sinx-f'(sinx)cos x

Is that correct?
The problem is I shoudl get
y'''(x)=f'''(sinx)cos^3x-3cosxsinxf''(sinx)-f'(sinx)cosx

where did I make the mistake?
Many thanks!!!
• Jan 14th 2013, 06:30 AM
Plato
Re: y=f(sinx)
Quote:

Originally Posted by Boo
Hello!

then
y'(x) should be f'(sinx)cosx
y''(x)= f''(sinx) cos x+f'(sinx)(-sinx)
where did I make the mistake?

$\displaystyle y''=f''(\sin(x))\cos^2(x)-\sin(x)f'(\cos(x))$
• Jan 15th 2013, 02:30 AM
Boo
Re: y=f(sinx)
Hello, Plato!
Thank You!
I understand how U got f''(sin(x))cos^2(x)
it is chain rule.

But how did U get -sin(x)f'(cosx)?

I tzhought it is now the product rule:
(f'x(x)cosx)) so we should have f1'(x)f2(x)+f1(x)f2'(x)???
Is it ok?
• Jan 15th 2013, 03:32 AM
ibdutt
Re: y=f(sinx)
• Jan 16th 2013, 03:49 AM
Boo
Re: y=f(sinx)
Thank You, Plato and Ibdutt!
You are life-saving!!!:)