Answer to Question #194251 in Calculus for Maureen

Question

Use the technique of differentiation to find dy over dx

Y=(xcube+7xsquared-8) (2x to the power -3 + x to the power -4)

Answer 1

$y=(x^3+7x^2-8)(2x^{-3}+x^{-4})\newline \text{differentiate w.r.t x,}\newline y'=(x^3+7x^2-8)(-6x^{-4}-4x^{-5})+(2x^{-3}+x^{-4})(3x^2+14x)\newline =-6x^{-1}-4x^{-2}-42x^{-2}-28x^{-3}+48x^{-4}+32x^{-5}+6x^{-1}+28x^{-2}+3x^{-2}+14x^{-3}\newline =-15x^{-2}-14x^{-3}+48x^{-4}+32x^{-5} \newline \newline$