1. ## differentiate trig

Differentiate

xsin²x

and

sinxcos²x

any help thanks

2. Originally Posted by gracey
Differentiate

xsin²x

and

sinxcos²x

any help thanks
Do you know product rule?

It says if f(x) and g(x) are two functions of x, then $\displaystyle (f(x) g(x))' = (f(x))'g(x) + f(x)(g(x))'$, where $\displaystyle (f(x))'$ stands for differentiationof $\displaystyle f(x)$.

Now choose $\displaystyle f(x) = x$and $\displaystyle g(x) = \sin x$ and apply the above rule. Do a similar substitution for the second problem too.

3. The product rule states that if a function, $\displaystyle f(x)$, is a product of two functions, $\displaystyle u$ and $\displaystyle v$, then $\displaystyle f'(x) = uv' + vu'$

For $\displaystyle f(x) = x \sin^{2}x$, let $\displaystyle u = x$ and $\displaystyle v = \sin^{2} x.$

It follows that $\displaystyle u' = 1$ and $\displaystyle v' = 2 \sin x \cos x$.

Then $\displaystyle f'(x) = x (2 \sin x \cos x) + \sin^{2}x (1)$

$\displaystyle = 2x \sin x \cos x + \sin^{2} x.$

$\displaystyle = \sin x (2x \cos x + \sin x).$

Following the same procedure for the second function, you get

$\displaystyle f'(x) = \sin x (2 \cos x \cdot -\sin x) + \cos^{2} x (\cos x)$

$\displaystyle = -2 \cos x \sin^2 x + \cos^{3} x$

$\displaystyle = -2 \cos x (1 - \cos^{2} x) + \cos^{3} x$

$\displaystyle = -2 \cos x + 2 \cos^{3} x + \cos^{3} x$

$\displaystyle = \cos x (-2 + 2 \cos^{2} x + \cos^{2} x)$

$\displaystyle = \cos x (3 \cos^{2} x - 2).$

or, if you instantly realise that $\displaystyle \sin x \cos^{2} x = \sin x (1 - \sin^{2} x) = \sin x - \sin^{3} x$, you do not need to use the product rule!

$\displaystyle f'(x) = \cos x - 3 \sin^{2} x \cos x$

$\displaystyle = \cos x - 3 \cos x (1 - \cos^{2} x)$

$\displaystyle = \cos x - 3 \cos x + 3 \cos^{3} x$

$\displaystyle = -2 \cos x + 3 \cos^{3} x$

$\displaystyle = \cos x (3 \cos^{2} x - 2)$.

I hope that helps.

ILoveMaths07.