Chapter 1.4, Problem 55E

To determine

To find: The slope of the curve at the given point.

Answer to Problem 55E

0.540.

Explanation of Solution

Given information:

The function is $y=\sin x$ at $x=1$ .

Concept used:

Slope of any function f at any point a is given by: $f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ .

Calculation:

Slope $y=\sin x$ at $x=1$ will be:

  $\begin{array}{c}{f}^{\prime }\left(1\right)=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}\\ =\underset{h\to 0}{\lim }\left(\frac{\sin \left(1+h\right)-\sin \left(1\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\sin \left(h\right)\cos \left(1\right)+\sin \left(1\right)\cos \left(h\right)-\sin \left(1\right)}{h}\right)\\ =\underset{h\to 0}{\lim }\left(\frac{\sin \left(h\right)\cos \left(1\right)}{h}\right)+\underset{h\to 0}{\lim }\left(\frac{\sin \left(1\right)\left(\cos \left(h\right)-1\right)}{h}\right)\\ =\cos \left(1\right)\underset{h\to 0}{\lim }\left(\frac{\sin \left(h\right)}{h}\right)+\sin \left(1\right)\underset{h\to 0}{\lim }\left(\frac{\left(\cos \left(h\right)-1\right)}{h}\right)\\ =\cos \left(1\right)\underset{h\to 0}{\lim }\left(\frac{\cos \left(h\right)}{1}\right)+\sin \left(1\right)\underset{h\to 0}{\lim }\left(\frac{\left(\sin \left(h\right)-0\right)}{1}\right)\left(\text{Apply L'Hopital rule}\right)\\ =\cos \left(1\right)+\left(0\right)\sin \left(1\right)\\ =\cos \left(1\right)\\ \approx 0.540\end{array}$

Conclusion:

The slope of the curve at the given point is $0.540$ .