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Find the derivative of the following function.

y=ln(9x^(2))
Answer:
y^(')=

Find the derivative of the following function. $\newline$ $y=\ln \left(9 x^{2}\right)$ $\newline$ Answer: $y^{\prime}=$

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Find derivatives of radical functions

Full solution

Q. Find the derivative of the following function.

\newline

y=\ln \left(9 x^{2}\right)

\newline

Answer:

y^{\prime}=

Identify function: Identify the function to differentiate. $\newline$ We have $y = \ln(9x^{2})$ . We need to find the derivative of $y$ with respect to $x$ , denoted as $y'$ .
Apply chain rule: Apply the chain rule for differentiation. $\newline$ The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. $\newline$ Here, the outer function is $\ln(u)$ and the inner function is $u = 9x^{2}$ .
Differentiate outer function: Differentiate the outer function with respect to the inner function. $\newline$ The derivative of $\ln(u)$ with respect to $u$ is $\frac{1}{u}$ . $\newline$ So, $\frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx}$ .
Differentiate inner function: Differentiate the inner function with respect to $x$ . The inner function is $u = 9x^{2}$ . The derivative of $9x^{2}$ with respect to $x$ is $18x$ . So, $\frac{du}{dx} = 18x$ .
Substitute derivatives: Substitute the derivatives into the chain rule formula. $\newline$ We have $\frac{d}{dx}[\ln(u)] = \frac{1}{u} \cdot \frac{du}{dx}$ . $\newline$ Substituting the derivatives from steps $3$ and $4$ , we get $\frac{d}{dx}[\ln(9x^{2})] = \frac{1}{9x^{2}} \cdot 18x$ .
Simplify expression: Simplify the expression. $\newline$ Simplify the derivative to get $y' = \frac{18x}{9x^{2}}$ . $\newline$ This simplifies further to $y' = \frac{2}{x}$ .

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