How do you use the product rule to differentiate #sqrt(1+3x^2 )lnx^2#?

Question

Christopher Allers · Accepted Answer

Solution

Assume the following. #u = sqrt(1+3x^2)# #v = ln x^2# #d(uv)= u .dv+v.du# #d(uv) = sqrt(1+3x^2) imes1/x^2 imes 2x + ln x^2 imes 1/(2sqrt(1+3x^2)) imes6x# # = (2sqrt(1+3x^2))/x+(3xlnx^2)/sqrt(1+3x^2)# On further reduction will lead to #(d(uv))/dx = (2+6x^2+3x^2ln x^2)/(xsqrt(1+3x^2))#

Samantha Adkison · Answer

undefined To use the product rule to differentiate $\sqrt{1 + 3x^2} \ln(x^2)$, you first identify the two functions being multiplied together: $\sqrt{1 + 3x^2}$ and $\ln(x^2)$. Then, apply the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Let $f(x) = \sqrt{1 + 3x^2}$ and $g(x) = \ln(x^2)$.

Using the product rule, the derivative of the given expression is:

$$f'(x) = \frac{d}{dx}\left(\sqrt{1 + 3x^2}\right) \ln(x^2) + \sqrt{1 + 3x^2} \frac{d}{dx}\left(\ln(x^2)\right)$$

Now, differentiate each function separately:

$$\frac{d}{dx}\left(\sqrt{1 + 3x^2}\right) = \frac{1}{2\sqrt{1 + 3x^2}} \cdot \frac{d}{dx}(1 + 3x^2)$$
$$= \frac{1}{2\sqrt{1 + 3x^2}} \cdot (0 + 6x) = \frac{3x}{\sqrt{1 + 3x^2}}$$

$$\frac{d}{dx}\left(\ln(x^2)\right) = \frac{1}{x^2} \cdot \frac{d}{dx}(x^2) = \frac{1}{x^2} \cdot 2x = \frac{2}{x}$$

Substitute these derivatives back into the product rule formula:

$$f'(x) = \left(\frac{3x}{\sqrt{1 + 3x^2}}\right) \ln(x^2) + \left(\sqrt{1 + 3x^2}\right) \left(\frac{2}{x}\right)$$

This expression is the derivative of the given function.