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Chapter 3: Problem 39

Find the derivative of the following functions. $$y=\frac{1}{\sec z \csc z}$$

Short Answer

Expert verified
Question: Find the derivative of the function $$y=\frac{1}{\sec z \csc z}$$. Answer: The derivative of the function $$y=\frac{1}{\sec z \csc z}$$ is $$y'=\cos^2 z - \sin^2 z$$.

Step by step solution

01

Identify the functions involved

The given function is $$y=\frac{1}{\sec z \csc z}$$. Write the trigonometric functions in terms of sine and cosine functions: $$y=\frac{1}{\frac{1}{\cos{z}}\frac{1}{\sin{z}}}$$ Simplified, $$y=\sin z \cos z$$ Now, we need to find the derivative $$y'=\frac{dy}{dz}$$.
02

Apply product rule

As we are dealing with the product of two functions (sin and cos), we will use the product rule. Recall that the product rule states: \((u \cdot v)'=u' \cdot v + u \cdot v'\). Here, let $$u = \sin z$$ and $$v = \cos z$$. To apply the product rule, we must first find the derivatives $$\frac{du}{dz}$$ and $$\frac{dv}{dz}$$. We know that $$\frac{d}{dz}(\sin z)=\cos z$$ and $$\frac{d}{dz}(\cos z)=-\sin z$$. Now proceed to apply the product rule to calculate the derivative: $$y'=\frac{dy}{dz} = (\sin z \cos z )' = \left( \frac{d}{dz}(\sin z)\right) \cos z + \sin z \left(\frac{d}{dz}(\cos z)\right)$$
03

Substitute the derivatives and simplify

Substitute the derivatives of sin z and cos z that we found earlier, and simplify: $$y'=\left( \cos z \right) \cos z + \sin z \left( -\sin z \right)$$ $$y'=\cos^2 z - \sin^2 z$$ Therefore, the derivative of the given function $$y=\frac{1}{\sec z \csc z}$$, is $$y'=\cos^2 z - \sin^2 z$$.

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