Use the quotient rule to find the derivative.
-$g(x)=\frac{x^{2}+5}{x^{2}+6 x}$
A) $g^{\prime}(x)=\frac{6 x^{2}-10 x-30}{x^{2}(x+6)^{2}}$
B) $g^{\prime}(x)=\frac{x^{4}+6 x^{3}+5 x^{2}+30 x}{x^{2}(x+6)^{2}}$
C) $g^{\prime}(x)=\frac{2 x^{3}-5 x^{2}-30 x}{x^{2}(x+6)^{2}}$
D) $g^{\prime}(x)=\frac{4 x^{3}+18 x^{2}+10 x+30}{x^{2}(x+6)^{2}}$

Question

Quizplus · Accepted Answer

The quotient rule states that the derivative of a quotient of two functions f(x) and g(x) is given by:$$h(x)=\frac{f(x)}{g(x)}$$$$h'(x)=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$$In this case, $$f(x)=x^2+5$$$$g(x)=x^2+6x$$So,$$f'(x)=2x$$$$g'(x)=2x+6$$Substituting these values into the quotient rule, we get:$$g'(x)=\frac{(x^2+6x)(2x)-(x^2+5)(2x+6)}{(x^2+6x)^2}$$Simplifying this expression, we get:$$g'(x)=\frac{2x^3+12x^2-2x^3-10x-6x^2-30}{(x^2+6x)^2}$$$$g'(x)=\frac{6x^2-10x-30}{(x^2+6x)^2}$$

Use the Quotient Rule to Find the Derivative g(x)=x2+5x2+6xg(x)=\frac{x^{2}+5}{x^{2}+6 x}g(x)=x2+6xx2+5​ A) g′(x)=6x2−10x−30x2(x+6)2g^{\prime}(x)=\frac{6 x^{2}-10 x-30}{x^{2}(x+6)^{2}}g′(x)=x2(x+6)26x2−10x−30​

Use the Quotient Rule to Find the Derivative $g(x)=\frac{x^{2}+5}{x^{2}+6 x}$ A) $g^{\prime}(x)=\frac{6 x^{2}-10 x-30}{x^{2}(x+6)^{2}}$