Question

section 2.6: chain rule (homework)
score: 110/170 answered: 11/17
progress saved done
question 12
0/10 pts 4 99 details
use the chain rule to find the derivative of 6\sqrt{3x^{4}+9x^{5}}
type your answer without fractional or negative exponents. use sqrt(x) for \sqrt{x}.

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:

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