Chapter 1: Problem 6
Find \(d^{2} y / d x^{2}\) $$ y=4 x^{2}+3 x-1 $$
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Chapter 1: Problem 6
Find \(d^{2} y / d x^{2}\) $$ y=4 x^{2}+3 x-1 $$
These are the key concepts you need to understand to accurately answer the question.
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Let \(f\) and \(g\) be differentiable over an open interval containing \(x=a\). If $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{0}{0} \quad \text { or } \quad \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\frac{\pm \infty}{\pm \infty} $$ and if \(\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right)\) exists, then $$ \lim _{x \rightarrow a}\left(\frac{f(x)}{g(x)}\right)=\lim _{x \rightarrow a}\left(\frac{f^{\prime}(x)}{g^{\prime}(x)}\right) $$ The forms \(0 / 0\) and \(\pm \infty / \pm \infty\) are said to be indeterminate. In such cases, the limit may exist, and l'Hôpital's Rule offers a way to find the limit using differentiation. For example, in Example 1 of Section \(1.1,\) we showed that $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=2 $$ Since, for \(x=1,\) we have \(\left(x^{2}-1\right)(x-1)=0 / 0,\) we differentiate the numerator and denominator separately, and reevaluate the limit: $$ \lim _{x \rightarrow 1}\left(\frac{x^{2}-1}{x-1}\right)=\lim _{x \rightarrow 1}\left(\frac{2 x}{1}\right)=2 $$ Use this method to find the following limits. Be sure to check that the initial substitution results in an indeterminate form. $$ \lim _{x \rightarrow 1}\left(\frac{x^{3}+2 x-3}{x^{2}-1}\right) $$
Graph fand \(f^{\prime}\) and then determine \(f^{\prime}(1) .\) $$ f(x)=\frac{4 x}{x^{2}+1} $$
For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=5 x^{2}-3 x+8 $$
Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\frac{x^{5}+x}{x^{2}} $$
Find dy/dx. Each function can be differentiated using the rules developed in this section, but some algebra may be required beforehand. $$ y=\frac{x^{5}-x^{3}}{x^{2}} $$
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