# What is the derivative of f(x)= 5 secx tanx?

Oct 29, 2016

You can say that...

$y = f \left(x\right)$

This means that:

$y = 5 \sec x \tan x$

And as a consequence....

$\frac{1}{5} \cdot y = \sec x \tan x$

From here you can use implicit differentiation on the left hand side of the equation, then use the product rule on the right hand side of the equation.

If you do this you should get:

$\frac{1}{5} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \sec x \left(2 {\sec}^{2} x - 1\right)$

Which means that:

$\frac{\mathrm{dy}}{\mathrm{dx}} = 5 \sec x \left(2 {\sec}^{2} x - 1\right)$

Therefore:

$f ' \left(x\right) = 5 \sec x \left(2 {\sec}^{2} x - 1\right)$