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Find the derivative of the function. $\newline$ $G(z)=(1-9z)^{2}\sqrt{z^{2}+1}$

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Find derivatives of radical functions

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Q. Find the derivative of the function.

\newline

G(z)=(1-9z)^{2}\sqrt{z^{2}+1}

Identify Components: Identify the components of the function $G(z)$ that will require the use of the product rule and the chain rule. $\newline$ $G(z) = (1-9z)^{2} \cdot \sqrt{z^{2}+1}$ $\newline$ We have two functions here: $u(z) = (1-9z)^{2}$ and $v(z) = \sqrt{z^{2}+1}$ . $\newline$ We will need to use the product rule, which states that $(uv)' = u'v + uv'$ .
Derivative of $u(z)$ : Find the derivative of the first function $u(z) = (1-9z)^{2}$ using the chain rule. $\newline$ Let's set a new function $h(z) = 1-9z$ , so $u(z) = h(z)^{2}$ . $\newline$ The derivative of $h(z)$ with respect to $z$ is $h'(z) = -9$ . $\newline$ Now, using the chain rule, the derivative of $u(z)$ with respect to $z$ is $u'(z) = 2 \cdot h(z)^{2-1} \cdot h'(z)$ . $\newline$ Substitute $h(z)$ back in to get $u(z) = (1-9z)^{2}$ $1$ .
Derivative of $v(z)$ : Find the derivative of the second function $v(z) = \sqrt{z^{2}+1}$ using the chain rule. $\newline$ Let's set a new function $i(z) = z^{2}+1$ , so $v(z) = \sqrt{i(z)}$ . $\newline$ The derivative of $i(z)$ with respect to $z$ is $i'(z) = 2z$ . $\newline$ Now, using the chain rule, the derivative of $v(z)$ with respect to $z$ is $v'(z) = (1/2) \cdot i(z)^{-1/2} \cdot i'(z)$ . $\newline$ Substitute $i(z)$ back in to get $v(z) = \sqrt{z^{2}+1}$ $1$ .
Apply Product Rule: Apply the product rule to find the derivative of $G(z)$ . Using the product rule $(uv)' = u'v + uv'$ , we have: $G'(z) = u'(z)v(z) + u(z)v'(z)$ . Substitute the derivatives we found earlier to get: $G'(z) = (-18 \times (1-9z)) \times \sqrt{z^{2}+1} + (1-9z)^{2} \times \left(\frac{z}{\sqrt{z^{2}+1}}\right)$ .
Simplify $G'(z)$ : Simplify the expression for $G'(z)$ . $G'(z) = -18 \times (1-9z) \times \sqrt{z^{2}+1} + (1-9z)^{2} \times \left(\frac{z}{\sqrt{z^{2}+1}}\right)$ .This is the simplified form of the derivative, and no further simplification is needed unless we want to factor common terms or expand the expression, which is not required for finding the derivative.

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