how do u find the derivative of (csc x)^7 -cos 4x???
You should have a list handy for the basic trig function derivatives
so if y = (csc x)^7 -cos 4x
dy/dx = 7(csc x)^6 ( -csc x)(cotx) + 4sin(4x)
etc
Thanks. My teacher told us we need to use the chain rule in these problems, and i'm having trouble figuring out when and where to use it.
I did use the chain rule in finding the derivative of
(csc x)^7
I know. I just realized that I messed up what goes where when I was doing it myself. My problem is applying it to other problems, esp when i also have to use the quotient or product rule.
To find the derivative of the expression, (csc x)^7 - cos 4x, we can apply the rules of differentiation.
First, let's break down the expression into two separate terms: (csc x)^7 and -cos 4x.
1. Derivative of (csc x)^7:
To find the derivative of (csc x)^7, we can rewrite it in terms of sine:
(csc x)^7 = (1/sin x)^7 = sin^(-7) x
To differentiate the term sin^(-7) x, we use the chain rule. The derivative of sin^(-7) x is given by:
d/dx[ sin^(-7) x ] = -7(sin^(-8) x) * cos x
= -7 cos x / sin^(8) x
2. Derivative of -cos 4x:
The derivative of -cos 4x is calculated using the chain rule. The derivative of -cos 4x can be written as:
d/dx[ -cos 4x ] = -1 * d/dx[ cos 4x ]
= -1 * (-sin 4x) = sin 4x
Now, to find the derivative of the original expression, we add the derivatives of both terms:
d/dx[ (csc x)^7 - cos 4x ] = -7 cos x / sin^(8) x + sin 4x
Therefore, the derivative of (csc x)^7 - cos 4x is -7 cos x / sin^(8) x + sin 4x.