How do you use the limit definition to find the derivative of #y = cscx#?

Question

Ezra Adams · Accepted Answer

The key parts are in the explanation section, below.

#(1/sin(x+h)-1/sinx)/h = (sinx-sin(x+h))/(hsin(x+h)sinx)# # = (sinx-sinxcos h-cosxsin h)/(hsin(x+h)sinx)# # = (sinx(1-cos h)/h - cosxsin h/h)*1/(sin(x+h)sinx)# Taking the limit gets us #(-cosx)/sin^2 x = -cscxcotx#

Isabella Ackerman · Answer

undefined To find the derivative of $y = \csc(x)$ using the limit definition of the derivative, we start by recalling that the derivative of a function $f(x)$ is given by:

$$f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}$$

For the function $y = \csc(x)$, we can rewrite it as $y = \frac{1}{\sin(x)}$. Now, applying the limit definition:

$$y'(x) = \lim_{h \to 0} \frac{\csc(x + h) - \csc(x)}{h}$$

Using the trigonometric identity $\csc(x) = \frac{1}{\sin(x)}$, we rewrite the expression:

$$y'(x) = \lim_{h \to 0} \frac{\frac{1}{\sin(x + h)} - \frac{1}{\sin(x)}}{h}$$

Next, we'll use the difference of squares formula for sine:

$$\sin(A + B) - \sin(A) = 2\cos\left(\frac{A + B}{2}\right)\sin\left(\frac{B}{2}\right)$$

Applying this identity, we get:

$$y'(x) = \lim_{h \to 0} \frac{\sin(x) - \sin(x + h)}{h\sin(x)\sin(x + h)}$$

Now, as $h$ approaches 0, we recognize that the expression becomes the derivative of $\sin(x)$ with respect to $x$, which is $\cos(x)$. So, the derivative of $y = \csc(x)$ is:

$$y'(x) = -\cot(x)\csc(x)$$