$\bf y=(2x-3)^5(2-x^4)^3\\\\
-------------------------------\\\\
\cfrac{dy}{dx}=\stackrel{\textit{product rule}}{[5(2x-3)^4\cdot 2(2-x^4)3]~~+~~[(2x-3)^5[3(2-x^4)^2(-4x^3)]]}
\\\\\\
\cfrac{dy}{dx}=10(2x-3)^4(2-x^4)^3~~-~~12x^3(2x-3)^5(2-x^4)^2
\\\\\\
\cfrac{dy}{dx}=\stackrel{\textit{common factor}}{2(2x-3)^4(2-x^4)^2}~[5(2-x^4)-6x^3(2x-3)]
\\\\\\
\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2~[10-5x^4-12x^4+18x^3]
\\\\\\
\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2(10-17x^4+18x^3)$

5 0

3 years ago

Help! What is the cost of 50 shares of stock priced at $15.87.

lozanna [386]

$793.50, All you do is multiply $15.87 by 50

3 0

3 years ago

Which expressions are equivalent to 4(3j+(-4))-9

Aleksandr [31]

4 • (3j - 4) - 9
 12j - 25

7 0

3 years ago