`y = ln|sec(x) + tan(x)|` Find the derivative of the function.

Tuesday, January 5, 2016

$\displaystyle{y}={\ln}{\left|{\sec{{x}}}+{\tan{{x}}}\right|}$

To take the derivative of this function, use the formula:

$\displaystyle{\left({\ln{{u}}}\right)}'=\frac{{1}}{{u}}\cdot{u}'$

Applying this formula, y' will be:

$\displaystyle{y}'=\frac{{1}}{{{\sec{{x}}}+{\tan{{x}}}}}\cdot{\left({\sec{{x}}}+{\tan{{x}}}\right)}'$

To get the derivative of the inner function, use the formulas:

$\displaystyle{\left({\sec{\theta}}\right)}'={\sec{\theta}}{\tan{\theta}}$

$\displaystyle{\left({\tan{\theta}}\right)}'={{\sec}}^{{2}}\theta$

So y' will become:

$\displaystyle{y}'=\frac{{1}}{{{\sec{{x}}}+{\tan{{x}}}}}\cdot{\left({\sec{{x}}}{\tan{{x}}}+{{\sec}}^{{2}}{x}\right)}$

Simplifying it, the derivative will be:

$\displaystyle{y}'=\frac{{{\sec{{x}}}{\tan{{x}}}+{{\sec}}^{{2}}{x}}}{{{\sec{{x}}}+{\tan{{x}}}}}$

$\displaystyle{y}'=\frac{{{\sec{{x}}}{\left({\tan{{x}}}+{\sec{{x}}}\right)}}}{{{\sec{{x}}}+{\tan{{x}}}}}$

$\displaystyle{y}'=\frac{{{\sec{{x}}}{\left({\sec{{x}}}+{\tan{{x}}}\right)}}}{{{\sec{{x}}}+{\tan{{x}}}}}$

$\displaystyle{y}'={\sec{{x}}}$

Therefore, the derivative of the given function is $\displaystyle{y}'={\sec{{x}}}$ .

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