How do I derive cos(4lnt) ??

Question

anonymous · Answer

My question is differentiate  $$t ^{-3}.\cos(4lnt)$$

I know it's product rule but I don't have a clue how to 2nd product..

anonymous · Answer

derive the 2nd product

amistre64 · Answer

its called a chain rule

amistre64 · Answer

essentially you have

f(4g(x)) derives to:  f'(4(g(x)) * 4g'(x)

anonymous · Answer

@amistre64 Sorry my internet connection went down. That seems fine. I got (4sin(lnt))/t

anonymous · Answer

Wolfram says otherwise, can you see where I went wrong.

anonymous · Answer

I let 4lnt = u

anonymous · Answer

d/dx cos u = sinu

amistre64 · Answer

t^(-3) *  cos(4lnt)

t^(-3)' *  cos(4lnt) + t^(-3) *  cos'(4lnt)

-3t^(-4) cos(4lnt) + t^(-3) * (-sin(4lnt))*4ln't

-3t^(-4) cos(4lnt) - t^(-3) * sin(4lnt)) * 4/t

amistre64 · Answer

derivatice of cos is -sin

amistre64 · Answer

$$-\frac{cos(4lnt)}{3t^4} - \frac{4sin(4lnt)}{t^3t^1}$$
$$-\frac{cos(4lnt)}{3t^4} - \frac{4sin(4lnt)}{t^4}$$

amistre64 · Answer

that 3 should be up top :)

amistre64 · Answer

$$-\frac{1}{t^4}(~3cos(4lnt)+4sin(4lnt)~)$$

amistre64 · Answer

i spose a +C is needed for correctness in the end

anonymous · Answer

-3t^(-4) cos(4lnt) - t^(-3) * sin(4lnt)) * 4/t

Just wondering where the minus on the sin go?

amistre64 · Answer

-3t^(-4) cos(4lnt)  -   t^(-3) * sin(4lnt)) * 4/t
                           ^^
                         the -sin part makes the whole thing go negative so it
                           switches the operator

anonymous · Answer

Oh you brought it out in front of the term.