Chapter 3: Problem 1
Use \(x=e^{y}\) to explain why \(\frac{d}{d x}(\ln x)=\frac{1}{x},\) for \(x > 0\).
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Chapter 3: Problem 1
Use \(x=e^{y}\) to explain why \(\frac{d}{d x}(\ln x)=\frac{1}{x},\) for \(x > 0\).
These are the key concepts you need to understand to accurately answer the question.
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The following limits equal the derivative of a function \(f\) at a point a. a. Find one possible \(f\) and \(a\) b. Evaluate the limit. $$\lim _{h \rightarrow 0} \frac{\sin \left(\frac{\pi}{6}+h\right)-\frac{1}{2}}{h}$$
Derivatives and inverse functions Find the slope of the curve \(y=f^{-1}(x)\) at (4,7) if the slope of the curve \(y=f(x)\) at (7,4) is \(\frac{2}{3}\)
Compute the derivative of the following functions. \(h(x)=\frac{(x+1)}{x^{2} e^{x}}\)
Let $$g(x)=\left\\{\begin{array}{cl} \frac{1-\cos x}{2 x} & \text { if } x \neq 0 \\ a & \text { if } x=0 \end{array}\right.$$ For what values of \(a\) is \(g\) continuous?
Proof of the Quotient Rule Let \(F=f / g\) be the quotient of two functions that are differentiable at \(x\) a. Use the definition of \(F^{\prime}\) to show that \(\frac{d}{d x}\left[\frac{f(x)}{g(x)}\right]=\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}\) b. Now add \(-f(x) g(x)+f(x) g(x)\) (which equals 0) to the numerator in the preceding limit to obtain $$\lim _{h \rightarrow 0} \frac{f(x+h) g(x)-f(x) g(x)+f(x) g(x)-f(x) g(x+h)}{h g(x+h) g(x)}$$ Use this limit to obtain the Quotient Rule. c. Explain why \(F^{\prime}=(f / g)^{\prime}\) exists, whenever \(g(x) \neq 0\)
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