Question

Find \(y^{\prime \prime}\) for the following functions. $$y=\sec x \csc x$$

Short answer

Answer: The second derivative of \(y = \sec(x) \csc(x)\) is \(y'' = \sec(x)\csc(x)(\tan(x) \cot(x) - \csc(x) + \sec(x) \tan(x) + \tan^2(x))\).

Step-by-step solution

Find the first derivative of the function using the product rule

The given function is: $$y = \sec(x) \csc(x)$$ To find the first derivative, use the product rule: $$(uv)' = u'v + uv'$$ Derivative of \(\sec(x)\) is \(\sec(x) \tan(x)\), and the derivative of \(\csc(x)\) is \(-\csc(x) \cot(x)\). Using the product rule, the first derivative is: $$y' = (\sec(x))(-\csc(x) \cot(x)) + (\csc(x))(\sec(x) \tan(x))$$

Simplify the expression of the first derivative

Expanding the expression from the previous step: $$y' = -\sec(x) \csc(x) \cot(x) + \sec(x) \tan(x) \csc(x)$$

Find the second derivative using the product rule

Now, find the second derivative by taking the derivative of the first derivative. Derivative of \(-\sec(x) \csc(x) \cot(x)\) is $$\sec(x) \tan(x) \csc(x) \cot(x) - \sec(x) \csc(x)^2$$ Derivative of \(\sec(x) \tan(x) \csc(x)\) is $$\sec(x)^2 \tan(x) \csc(x) + \sec(x) \tan^2(x) \csc(x) $$ Combining the derivatives, we get the second derivative: $$y'' = \sec(x) \tan(x) \csc(x) \cot(x) - \sec(x) \csc(x)^2 + \sec(x)^2 \tan(x) \csc(x) + \sec(x) \tan^2(x) \csc(x)$$

Simplify the expression of the second derivative

The second derivative can be simplified by combining like terms and factoring out the common term \(\sec(x)\csc(x)\): $$y'' = \sec(x)\csc(x)(\tan(x) \cot(x) - \csc(x) + \sec(x) \tan(x) + \tan^2(x))$$ We now have both the first and second derivatives of the function.