Problem 55
Question
Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ f(x)=\frac{1}{\left(x^{2}-3 x\right)^{2}} $$
Step-by-Step Solution
Verified Answer
The derivative of the given function is \(f'(x)= \frac{2*(2x-3)}{(x^{2}-3x)^{3}}\)
1Step 1: Identify the Inner and Outer Functions
A key step in solving this problem is determining the inner and outer functions. Here, the outer function is \(1/u^{2}\) and the inner function is \(u=x^{2}-3x\).
2Step 2: Apply Chain Rule
Apply the Chain Rule, which states that the derivative of a composite function is the derivative of the outer function, times the derivative of the inner function. Here, the derivative of the outer function is \(-2/u^{3}\) and the derivative of the inner function (found by applying the Power Rule) is \(2x-3\).
3Step 3: Substitute the Derivatives into Chain Rule
Substitute the derived expressions into the Chain Rule equation. We have \((-2/u^{3})*(2x-3)\). Now, substitute \(u=x^{2}-3x\) back into the equation to get the final answer.
4Step 4: Simplify the Result
Simplify the results to get the final expression. The derivative \[f'(x)= \frac{2*(2x-3)}{(x^{2}-3x)^{3}}\]
Other exercises in this chapter
Problem 54
Describe the \(x\) -values at which the function is differentiable. Explain your reasoning. $$ y=x^{2 / 5} $$
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find the limit $$ \lim _{x \rightarrow 1} f(s), \text { where } f(s)=\left\\{\begin{array}{ll} s, & s \leq 1 \\ 1-s, & s>1 \end{array}\right. $$
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Use the demand function to find the rate of change in the demand \(x\) for the given price \(p\). $$ x=275\left(1-\frac{3 p}{5 p+1}\right), p=\$ 4 $$
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Determine the point(s), if any, at which the graph of the function has a horizontal tangent line. $$ y=\frac{1}{2} x^{2}+5 x $$
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