Question

y=(2x^2-3)^5
y=(4+7x^3)^6
y=(2-3x^2)^8
y=(1-5x^5)^5

Answer 1

$y=(2x^2-3)^5\; ,\; \; y'=5(2x^2-3)^4\cdot 4x\\\\y=(4+7x^3)^6\; ,\; \; y'=6(4+7x^3)^5\cdot 21x^2\\\\y=(2-3x^2)^8\; ,\; \; y'=8(2-3x^2)^7\cdot (-6x)=-48x(2-3x^2)^7\\\\y=(1-5x^5)^5\; ,\; \; y'=5(1-5x^5)^4\cdot (-25x^4)=-125x^4(1-5x^5)^4$