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.
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.