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Chapter 2: Problem 16

Find the derivative by the limit process. \(f(x)=x^{3}+x^{2}\)

Short Answer

Expert verified
The derivative of \(f(x) = x^{3} + x^{2}\) by the limit process is \(f'(x) = 3x^{2} + 2x\).

Step by step solution

01

Define Limit

The definition of the derivative of a function \(f(x)\) at a point \(x\) is: \(f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\)
02

Apply the Limit to the Function

By applying the definition of the limit to the function \(f(x) = x^{3} + x^{2}\), we get: \[f'(x) = \lim_{h \to 0} \frac{((x+h)^{3} + (x+h)^{2}) - (x^{3} + x^{2})}{h}\]
03

Simplify

By simplifying the expression in the numerator, we have: \[f'(x) = \lim_{h \to 0} \frac{(x^{3} + 3x^{2}h + 3xh^{2} + h^{3} + x^{2} + 2xh + h^{2}) - (x^{3} + x^{2})}{h}\] Which simplifies to \[f'(x) = \lim_{h \to 0} \frac{3x^{2}h + 3xh^{2} + h^{3} + 2xh + h^{2}}{h} \] And further simplification by cancelling out \(h\) from all terms in the numerator gives us: \[f'(x) = 3x^{2} + 3x + h + 2x + h \] Now, as \( h \to 0 \), the terms with \(h\) in the equation above become zero.

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In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{f(x)=\tan ^{2} x} \quad \frac{\text { Point }}{\left(\frac{\pi}{4}, 1\right)}\)

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