Question

If y=(x^2-4)^5(3x+4)^4

Answer 1

Answer:

$y'=2(3x+4)^3(x^2-4)^4(21x^2+20x-24)$

Step-by-step explanation:

$y=(x^2-4)^5(3x+4)^4$

$y'=[\frac{d}{dx}(x^2-4)^5](3x+4)^4+(x^2-4)^5[\frac{d}{dx}(3x+4)^4]$

$y'=10x(x^2-4)^4(3x+4)^4+(x^2-4)^5(12(3x+4)^3)$

$y'=10x(x^2-4)^4(3x+4)^4+12(x^2-4)^5(3x+4)^3$

$y'=2(3x+4)^3(x^2-4)^4(21x^2+20x-24)$