did i use the chain rule correctly?
y=(8x^4-5x^2+1)^4
d(f(x)/dx =d/dx ((8x^4-5x^2+1)^4)
=4*(8x^4-5x^2+1)^3*d/dx(8x^4-5x^2+1)
=4*(8*d/dx(x^4)-5*d/dx(x^2))*(8x^4-5x^2+1)^3
=4*(8*4x^3-5*2x)*(8x^4-5x^2+1)^3
=4*(32x^3-10x)*(8x^4-5x^2+1)^3
3 answers
Yes.
(8x^4-5x^2+1)^4
=4(8x^4 - 5x^2 + 1)*d/dx( 8x^4-5x^2+1)
= 4(8x^4 - 5x^2 + 1)^3*(32x^3-10x^2)
=4(8x^4 - 5x^2 + 1)*d/dx( 8x^4-5x^2+1)
= 4(8x^4 - 5x^2 + 1)^3*(32x^3-10x^2)
All you do is differentiate the outter exponent first. then multiply that by the derivative of the inside parenthesis