Problem of the Week

Updated at Dec 7, 2015 3:58 PM

This week we have another calculus problem:

How would you differentiate $$\sec{x}\ln{x}$$?

Let's start!

$\frac{d}{dx} \sec{x}\ln{x}$

 1 Use Product Rule to find the derivative of $$\sec{x}\ln{x}$$. The product rule states that $$(fg)'=f'g+fg'$$.$(\frac{d}{dx} \sec{x})\ln{x}+\sec{x}(\frac{d}{dx} \ln{x})$2 Use Trigonometric Differentiation: the derivative of $$\sec{x}$$ is $$\sec{x}\tan{x}$$.$\sec{x}\tan{x}\ln{x}+\sec{x}(\frac{d}{dx} \ln{x})$3 The derivative of $$\ln{x}$$ is $$\frac{1}{x}$$.$\sec{x}\tan{x}\ln{x}+\frac{\sec{x}}{x}$Donesec(x)*tan(x)*ln(x)+sec(x)/x