Question

Differentiate t^4 In(8cost)

Answer 1

t^3 (4 ln(cos8t) - t tant)

Explanation:

Using the Product Rule:

dy/dt = t^4 * d(ln(8cost) / dt + ln(8cost) * d(t^4)/dt

= t^4 * 1/ (8cost) * (-8sint) + 4t^3 ln(8cost)

= -8t^4 sint / 8 cost + 4t^3 ln(8cost)

= -t^4 tan t + 4t^3 ln(8cost)

= t^3 (4 ln(cos8t) - t tant)

Answer 2

⇒It is way more appropriate if I use the product rule. That states that:

⇒f(x)g(x)=f'(x)g(x)+f(x)g'(x)

`t^(4) In(8cos(t))\\=4t^(3)In(8cos(t))+t^(4) (1)/(8cos(t)) *(0cos(t)+8*(-sin(t))*1)\\=4t^(3)In(8cos(t))+(t^(4)-8sin(t))/(8cos(t))`

Note:

Given F(x)=In(x)

⇒
`F'(x)=(1)/(x)`

Goodluck