Question

if f(x)= cosxsin3x, then f'(pi/6) =

Answer 1

can you take the derivative of the expression?

Answer 2

\(\large\color{blue}{ f(x)=\cos(x) \sin^3x }\)
\(\large\color{blue}{ f(x)=\cos(x) (\sin x )^3 }\)

Answer 3

Use the product rule, and for the (sin x)^3 do not forget to use the chain.

Answer 4

\(\large\color{blue}{ \frac{\LARGE{dy}}{\LARGE{dx}}=\frac{\LARGE{d}}{\LARGE{dx}}\cos(x) (\sin x )^3+\cos(x)\frac{\LARGE{d}}{\LARGE{dx}} (\sin x)^3 }\)

Answer 5

\(\large\color{blue}{ \frac{\LARGE{dy}}{\LARGE{dx}}=(-\sin~x)(\sin x )^3+\cos(x)\frac{\LARGE{d}}{\LARGE{dx}} (\sin x)^3 }\)

\(\large\color{blue}{ \frac{\LARGE{dy}}{\LARGE{dx}}=(-\sin~x)(\sin x )^3+\cos(x)3(\sin x)^2\times \frac{\LARGE{d}}{\LARGE{dx}}\sin(x) }\)

\(\large\color{blue}{ \frac{\LARGE{dy}}{\LARGE{dx}}=(-\sin~x)(\sin x )^3+\cos(x)3(\sin x)^2\times(\cos~x) }\)

Answer 6

\(\large\color{blue}{ \frac{\LARGE{dy}}{\LARGE{dx}}=-\sin^4x+3\cos^2(x)\sin ^2x }\)

Answer 7

is it f'(x)=cosx(sin3x)(cos3x)(3)+sin3x(-sinx)

Answer 8

\(\large\color{blue}{ \frac{\LARGE{dy}}{\LARGE{dx}}=-\sin^4x+3(1-\sin^2x)\sin ^2x }\)

Answer 9

\(\large\color{blue}{ \frac{\LARGE{dy}}{\LARGE{dx}}=-\sin^4x+3(\sin ^2x-\sin^4x)\ }\)

\(\large\color{blue}{ \frac{\LARGE{dy}}{\LARGE{dx}}=-\sin^4x+3\sin ^2x-3\sin^4x\ }\)

\(\large\color{blue}{ \frac{\LARGE{dy}}{\LARGE{dx}}=-4\sin^4x+3\sin ^2x\ }\)

Answer 10

this is the most simplified derivative expression, now plug in pi/6 for x

Answer 11

sin(30) is 1/2, and you know this from a 30-60-90 triangle.
So,

\(\large\color{blue}{ f'(pi/6)=-4(1/2)^4+3(1/2)^2\ }\)

Answer 12

all you need is to simplify this expression

Answer 13

i dont understand how you got dy/dx = -4sin^4x + 3sin^2x

Answer 14

I would suggest to go over my comments, because I was quite clear...

Answer 15

ok thanks