Thread: Derivative of 3cos^5x.

hi, my first post so pls be gentle!
HNC maths question driving me mad as must be doing something stupidly wrong.
3cos^5x is the question and result should be -15sinx cos^4x
i must be doing something fundamentally wrong as i cant match the answer given. any help greatly appreciated.

Originally Posted by neilj

hi, my first post so pls be gentle!
HNC maths question driving me mad as must be doing something stupidly wrong.
3cos^5x is the question and result should be -15sinx cos^4x
i must be doing something fundamentally wrong as i cant match the answer given. any help greatly appreciated.

derivative of $\displaystyle 3\cos^5{x} = -15\sin{x}cos^4{x}$

Just in case a picture helps...



... where



... is the chain rule. Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).
_________________________________________
Don't integrate - balloontegrate!

Balloon Calculus; standard integrals, derivatives and methods

Balloon Calculus Drawing with LaTeX and Asymptote!

Originally Posted by neilj

hi, my first post so pls be gentle!
HNC maths question driving me mad as must be doing something stupidly wrong.
3cos^5x is the question and result should be -15sinx cos^4x
i must be doing something fundamentally wrong as i cant match the answer given. any help greatly appreciated.

Put simpler still...

$\displaystyle \frac{dy}{dx} = \frac{du}{dx}\cdot\frac{dy}{du}$.


If $\displaystyle y = 3\cos^5{x}$

Let $\displaystyle u = \cos{x}$ so that $\displaystyle y = 3u^5$.


$\displaystyle \frac{du}{dx} = -\sin{x}$


$\displaystyle \frac{dy}{du} = 15u^4 = 15\cos^4{x}$.


Therefore $\displaystyle \frac{dy}{dx} = -15\sin{x}\cos^4{x}$.

Originally Posted by Prove It

Put simpler still...

$\displaystyle \frac{dy}{dx} = \frac{du}{dx}\cdot\frac{dy}{du}$.


If $\displaystyle y = 3\cos^5{x}$

Let $\displaystyle u = \cos{x}$ so that $\displaystyle y = 3u^5$.


$\displaystyle \frac{du}{dx} = -\sin{x}$


$\displaystyle \frac{dy}{du} = 15u^4 = 15\cos^4{x}$.


Therefore $\displaystyle \frac{dy}{dx} = -15\sin{x}\cos^4{x}$.

thanks for that. must have tried every variation bar that one!! splitting out into u and y parts was my downfall.