91Ó°ÊÓ

Find \(d y\) $$y=\sec ^{-1}\left(e^{-x}\right)$$

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$6 x^{2}+3 x y+2 y^{2}+17 y-6=0, \quad(-1,0)$$

Find \(d y\) $$y=e^{\ln x} \sqrt[1]{x^{2}+1}$$

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\sqrt{s^{2}-1}-\sec ^{-1} s$$

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta\) as appropriate. $$y=\ln \left(\frac{\sqrt{\sin \theta \cos \theta}}{1+2 \ln \theta}\right)$$

Find the derivatives of the functions. $$y=\left(9 x^{2}-6 x+2\right) e^{x^{3}}$$

Graph the curves over the given intervals, together with their tangent lines at the given values of \(x .\) Label each curve and tangent line with its equation. $$\begin{aligned}&y=1+\cos x, \quad-3 \pi / 2 \leq x \leq 2 \pi\\\&x=-\pi / 3,3 \pi / 2\end{aligned}$$

We say that a continuous curve \(y=f(x)\) has a vertical tangent line at the point where \(x=x_{0}\) if the limit of the difference quotient is \(\infty\) or - \(\infty .\) For example, \(y=x^{1 / 3}\) has a vertical tangent line at \(x=0\) (see accompanying figure): $$ \begin{aligned} \lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} &=\lim _{h \rightarrow 0} \frac{h^{1 / 3}-0}{h} \\ &=\lim _{h \rightarrow 0} \frac{1}{h^{2 / 3}}=\infty \end{aligned} $$ However, \(y=x^{2 / 3}\) has no vertical tangent line at \(x=0\) (see next figure): $$ \begin{aligned} \lim _{h \rightarrow 0} \frac{g(0+h)-g(0)}{h} &=\lim _{h \rightarrow 0} \frac{h^{2 / 3}-0}{h} \\ &=\lim _{h \rightarrow 0} \frac{1}{h^{1 / 3}} \end{aligned} $$ does not exist, because the limit is \(\infty\) from the right and \(-\infty\) from the Ieft. Does the graph of $$ U(x)=\left\\{\begin{array}{ll} 0, & x<0 \\ 1, & x \geq 0 \end{array}\right. $$ have a vertical tangent line at the point (0,1)\(?\) Give reasons for your answer

Verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point. $$x^{2}-\sqrt{3} x y+2 y^{2}=5, \quad(\sqrt{3}, 2)$$

Find the derivatives of the functions in Exercises \(17-40 .\) $$y=\sqrt[3]{x^{96}}+2 e^{1.3}$$