How do I find #f'(x)# for #f(x)=x^2*10^(2x)# ?

1 Answer
Sep 20, 2014

The derivative of #f(x)=x^2cdot10^{2x}# is

#f'(x)=2x(1+xln10)10^{2x}#

Let us look at some details.

We need the following tools in your toolbox.

  1. Power Rule: #(x^n)'=nx^{n-1}#

  2. Exponential Rule: #(b^x)'=(lnb)b^x#

  3. Product Rule: #[f(x)cdot g(x)]'=f'(x)cdot g(x)+f(x)cdot g'(x)#

  4. Chain Rule: #[f(g(x))]'=f'(g(x))cdot g'(x)#

Let us find #(10^{2x})'# first.

By Chain Rule and Exponential Rule,

#(10^{2x})'=(ln10)10^{2x}cdot(2x)'=2(ln10)10^{2x}#

Now, we can find #f'(x)#.

By Product Rule,

#f'(x)=(x^2)'cdot10^{2x}+x^2cdot(10^{2x})'#

by Power Rule and the derivative we found above,

#=2xcdot 10^{2x}+x^2cdot2(ln10)10^{2x}#

by factoring out #2x10^{2x}#,

#=2x(1+xln10)10^{2x}#