section 2.6: chain rule (homework)\nscore: 110/170 answered: 11/17\nprogress saved done\nquestion 12\n0/10…

Explanation:

Step1: Rewrite the function

Let (y = 6\sqrt{3x^{4}+9x^{5}}=6(3x^{4}+9x^{5})^{\frac{1}{2}}).

Step2: Apply the chain - rule

The chain - rule states that if (y = f(g(x))), then (y^\prime=f^\prime(g(x))\cdot g^\prime(x)). Let (u = 3x^{4}+9x^{5}), so (y = 6u^{\frac{1}{2}}). First, find (\frac{dy}{du}): (\frac{dy}{du}=6\times\frac{1}{2}u^{-\frac{1}{2}} = 3u^{-\frac{1}{2}}=\frac{3}{\sqrt{u}}). Then find (\frac{du}{dx}): (\frac{du}{dx}=\frac{d}{dx}(3x^{4}+9x^{5})=3\times4x^{3}+9\times5x^{4}=12x^{3}+45x^{4}).

Step3: Calculate (\frac{dy}{dx})

By the chain - rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). Substitute (u = 3x^{4}+9x^{5}) back in: (\frac{dy}{dx}=\frac{3}{\sqrt{3x^{4}+9x^{5}}}\times(12x^{3}+45x^{4})=\frac{3(12x^{3}+45x^{4})}{\sqrt{3x^{4}+9x^{5}}}). Multiply the numerator and denominator by (\sqrt{3x^{4}+9x^{5}}) to get rid of the square - root in the denominator: (\frac{dy}{dx}=\frac{3(12x^{3}+45x^{4})\sqrt{3x^{4}+9x^{5}}}{3x^{4}+9x^{5}}=\frac{(36x^{3}+135x^{4})\sqrt{3x^{4}+9x^{5}}}{3x^{4}+9x^{5}}).

Answer:

(\frac{(36x^{3}+135x^{4})\sqrt{3x^{4}+9x^{5}}}{3x^{4}+9x^{5}})