How do you find the derivative of Inverse trig function #y = (sin(3x) + cot(x^3))^8#?

1 Answer
Stefan V.
Aug 21, 2015

#y^' = 24 * [sin(3x) + cot(x^3)]^7 * [cos(3x) - x^2csc^2(x^3)]#

Explanation:

From the looks of it, you're going to have to use the chain rule three times to differentiate this function.

Before you get started, keep in mind that

#d/dx(sinx) = cosx" "# and #" "d/dx(cotx) = -csc^2x#

So, use the chain rule first for #u^8#, with #u = sin(3x) + cot(x^3)#

#d/dx(y) = d/(du)(u^8) * d/dx(u)#

#y^' = 8u^7 * d/dx(sin(3x) + cot(x^3))#

You can differentiate this function, #d/dx(sin(3x) + cot(x^3))# by using the chain rule twice, once for #sint#, with #t = 3x#, and once more for #cotv#, with #v = x^3#.

This means that you can write

#d/dx(sint) = d/(dt)sint * d/dx(t)#

#d/dx(sint) = cost * d/dx(3x)#

#d/dx(sin(3x)) = cos(3x) * 3#

and

#d/dx(cotv) = d/(dv)cotv * d/dx(v)#

#d/dx(cotv) = -csc^2v * d/dx(x^3)#

#d/dx(cot(x^3)) = -csc^2(x^3) * 3x^2#

Plug these derivatives into the calculation for #y^'# to get

#y^' = 8u^7 * [3cos(3x) - 3x^2csc^2(x^3)]#

#y^' = 8[sin(3x) + cot(x^3)]^7 * 3 * [cos(3x) - x^2csc^2(x^3)]#

.#y^' = color(green)(24 * [sin(3x) + cot(x^3)]^7 * [cos(3x) - x^2csc^2(x^3)])#

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