Find the derivative of $y=\frac{1}{\left(4 x^{3}-6 x+8\right)^{2}}$.
A) $\frac{d y}{d x}=\frac{12 x^{2}-6}{\left(4 x^{3}-6 x+8\right)^{3}}$
B) $\frac{d y}{d x}=\frac{-12\left(2 x^{2}-1\right)}{\left(4 x^{3}-6 x+8\right)^{4}}$
C) $\frac{d y}{d x}=\frac{12 x^{2}-6}{\left(4 x^{3}-6 x+8\right)^{4}}$
D) $\frac{d y}{d x}=\frac{-12\left(2 x^{2}-1\right)}{\left(4 x^{3}-6 x+8\right)^{3}}$

Question

Quizplus · Accepted Answer

The derivative is found using the chain rule and the power rule. The function is of the form $y = u^{-2}$ where $u = (4x^3 - 6x + 8)$. The derivative is $-2u^{-3} \cdot u'$, where $u' = 12x^2 - 6$. Substituting back, we get $-2(4x^3 - 6x + 8)^{-3} \cdot (12x^2 - 6)$, which simplifies to $\frac{-12(2x^2 - 1)}{(4x^3 - 6x + 8)^3}$.

Find the Derivative Of y=1(4x3−6x+8)2y=\frac{1}{\left(4 x^{3}-6 x+8\right)^{2}}y=(4x3−6x+8)21​ A) dydx=12x2−6(4x3−6x+8)3\frac{d y}{d x}=\frac{12 x^{2}-6}{\left(4 x^{3}-6 x+8\right)^{3}}dxdy​=(4x3−6x+8)312x2−6​

Find the Derivative Of $y=\frac{1}{\left(4 x^{3}-6 x+8\right)^{2}}$
A) $\frac{d y}{d x}=\frac{12 x^{2}-6}{\left(4 x^{3}-6 x+8\right)^{3}}$