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

Let us look at some details.

We need the following tools in your toolbox.

By Chain Rule and Exponential Rule,

By Product Rule,

by Power Rule and the derivative we found above,

By signing up, you agree to our Terms of Service and Privacy Policy

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)).

By signing up, you agree to our Terms of Service and Privacy Policy

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)).

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7