# Thread: differentiate trig

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 $(f(x) g(x))' = (f(x))'g(x) + f(x)(g(x))'$, where $(f(x))'$ stands for differentiationof $f(x)$.

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

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

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

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

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

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

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

Following the same procedure for the second function, you get

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

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

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

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

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

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

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

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

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

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

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

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

I hope that helps.

ILoveMaths07.