Differential help please

[val1]

Please can you check the following differentials?

Differentiate with respect to x.
a) $\displaystyle y = \cos ^3 x\tan 2x^2$


This is what I got...

$\displaystyle \frac{{dy}}{{dx}} = - 3\cos ^2 x\sin x.4\sec ^2 2x$


b)$\displaystyle y = \left( {\frac{{\tan ^2 2x}}{2} - x} \right)\sin x^3$

My workings so far...

\(\displaystyle \begin{array}{l}
y = \left( {\frac{{\tan ^2 2x}}{2} - x} \right)\sin x^3 = \left( {\frac{1}{2}\tan ^2 2x - x} \right)\sin x^3 \\
\frac{{dy}}{{dx}} = \left( {\frac{1}{4}\tan 2x\sec ^2 2x - x} \right)3\sin x^2 \cos x \\ \end{array}\)

Is this right so far?
How would I go on to simplify further?

Thanks for your help
:?

 

[skeeter]

have you heard of the product rule for derivatives?

y = cos³(x)tan(2x²)

y = u*v

dy/dx = u*v' + v*u'

where

u = cos³(x)
u' = -3cos²(x)sin(x)
v = tan(2x²)
v' = 4x*sec²(2x²)

 

[val1]

Thanks for the reply.

I will try the product rule for the first part of the question, but what do you think I should do with part b)?

b)$\displaystyle y = \left( {\frac{{\tan ^2 2x}}{2} - x} \right)\sin x^3$

My workings so far...

\(\displaystyle \begin{array}{l}
y = \left( {\frac{{\tan ^2 2x}}{2} - x} \right)\sin x^3 = \left( {\frac{1}{2}\tan ^2 2x - x} \right)\sin x^3 \\
\frac{{dy}}{{dx}} = \left( {\frac{1}{4}\tan 2x\sec ^2 2x - x} \right)3\sin x^2 \cos x \\ \end{array}\)

I don't think the product rule can help me with this one?

Thanks

 

[pka]

I don't think the product rule can help me with this one?

Why do you think that?
It is a product!

 

[val1]

Hi again

Thanks for pointing out it's a product.

Can you tell me if I'm on the right lines with this...?

\(\displaystyle \begin{array}{l}
u = \frac{1}{2}\tan ^2 2x - x \\
u' = - 2\sec ^2 2x - 1 \\
v = \sin x^3 \\
v' = 3\sin x^2 \cos x^3 \\
\end{array}\)

Thank you. :?

 

[skeeter]

almost ...

u = (1/2)tan^2(2x) - x
u' = 2tan(2x)sec^2(2x) - 1

v = sin(x^3)
v' = 3x^2*cos(x^3)