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

Question

Ezra Adams · Accepted Answer

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.

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

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

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

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}#

William Abdalla · Answer

undefined To find $f'(x)$ for $f(x) = x^2 \cdot 10^{2x}$, use the product rule and the chain rule. The derivative is:

$$f'(x) = 2x \cdot 10^{2x} + x^2 \cdot 10^{2x} \cdot \ln(10) \cdot 2$$

$$= 2x \cdot 10^{2x} + 2x^2 \cdot 10^{2x} \cdot \ln(10)$$

So, $f'(x) = 2x \cdot 10^{2x} + 2x^2 \cdot 10^{2x} \cdot \ln(10)$.

Evelyn Agard · Answer

undefined To find $f'(x)$ for $f(x) = x^2 \cdot 10^{2x}$, you can use the product rule of differentiation.

The product rule states that if you have two functions, $u(x)$ and $v(x)$, then the derivative of their product is given by $u'(x)v(x) + u(x)v'(x)$.

For $f(x) = x^2 \cdot 10^{2x}$, let $u(x) = x^2$ and $v(x) = 10^{2x}$.

Now, apply the product rule:

$$f'(x) = u'(x)v(x) + u(x)v'(x)$$

$$= (2x)(10^{2x}) + (x^2)(2\cdot10^{2x}\ln(10))$$

$$= 2x \cdot 10^{2x} + 2x^2 \cdot 10^{2x}\ln(10)$$

Therefore, $f'(x) = 2x \cdot 10^{2x} + 2x^2 \cdot 10^{2x}\ln(10)$.