Chapter 2: Problem 31
31-38. Find the indicated derivatives. If \(f(x)=x^{5}\), find \(f^{\prime}(-2)\).
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Chapter 2: Problem 31
31-38. Find the indicated derivatives. If \(f(x)=x^{5}\), find \(f^{\prime}(-2)\).
These are the key concepts you need to understand to accurately answer the question.
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The Chain Rule (page 155 ) states that the derivative of \(f(g(x))\) is \(f^{\prime}(g(x)) \cdot g^{\prime}(x)\). Use Carathéodory's definition of the derivative to prove the Chain Rule by giving reasons for the following steps. a. Since \(g\) is differentiable at \(x\), there is a function \(G\) that is continuous at 0 and such that \(g(x+h)-g(x)=G(h) \cdot h\), and \(G(0)=g^{\prime}(x)\). b. Since \(f\) is differentiable at \(g(x)\), there is a function \(F\) that is continuous at 0 and such that \(f(g(x)+h)-f(g(x))=F(h) \cdot h\), and \(F(0)=f^{\prime}(g(x))\) c. For the function \(f(g(x))\) we have \(f(g(x+h))-f(g(x))\) \(\quad=f(g(x)+g(x+h)-g(x))-f(g(x))\) \(\quad=f(g(x)+(g(x+h)-g(x)))-f(g(x))\) \(=F(g(x+h)-g(x)) \cdot(g(x+h)-g(x))\) \(=F(g(x+h)-g(x)) \cdot G(h) \cdot h\) Therefore, the derivative of \(f(g(x))\) is \(\begin{aligned} F(g(x+0)-g(x)) \cdot G(0) &=F(0) \cdot G(0) \\ &=f^{\prime}(g(x)) \cdot g^{\prime}(x) \end{aligned}\) as was to be proved.
True or False: If \(f(2)=5\), then \(\lim _{x \rightarrow 2} f(x)=5\).
True or False: \(\frac{d}{d x} f(5 x)=5 \cdot f^{\prime}(5 x)\)
Find \(f^{\prime}(x)\) by using the definition of the derivative. [Hint: See Example 4.] $$ \text { } f(x)=\frac{x}{2} $$
For each function: a. Find \(f^{\prime}(x)\) using the definition of the derivative. b. Explain, by considering the original function, why the derivative is a constant. $$ \begin{array}{l} f(x)=m x+b\\\ (m \text { and } b \text { are constants) } \end{array} $$
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