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How do you find the first and second derivative of # (ln(x^(2)+3))^(3)#?

Answer 1

Charles Arocha

#f'(x)=(6xln^2(x^2+3))/(x^2+3)#

#f''(x)=((-6x^2+18)ln^2(x^2+3) + 24x ln(x^2+3))/(x^2+3)^2#

The first derivative is:

#f'(x)=3ln^2(x^2+3)*1/(x^2+3)*2x#

#=(6xln^2(x^2+3))/(x^2+3)#

The second derivative is:

#f''(x)=((6ln^2(x^2+3) + 6x*2 ln(x^2+3)* 1/(x^2+3) * 2x)*(x^2+3)-6xln^2(x^2+3)*2x)/(x^2+3)^2#

#=((6ln^2(x^2+3) + (24x ln(x^2+3))/(x^2+3))*(x^2+3)-12x^2ln^2(x^2+3))/(x^2+3)^2#

#=(((6(x^2+3)ln^2(x^2+3) + 24x ln(x^2+3))/cancel(x^2+3))*cancel((x^2+3))-12x^2ln^2(x^2+3))/(x^2+3)^2#

#=(6(x^2+3)ln^2(x^2+3) + 24x ln(x^2+3)-12x^2ln^2(x^2+3))/(x^2+3)^2#

#=((6x^2+18-12x^2)ln^2(x^2+3) + 24x ln(x^2+3))/(x^2+3)^2#

#=((-6x^2+18)ln^2(x^2+3) + 24x ln(x^2+3))/(x^2+3)^2#

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Answer 2

Elijah Abram

To find the first derivative of the function $(ln(x^{2}+3))^{3}$ , apply the chain rule. The derivative of the outer function, $u^{3}$ , with respect to $u$ , is $3u^{2}$ , where $u = ln(x^{2}+3)$ . Then, the derivative of $ln(x^{2}+3)$ with respect to $x$ is $\frac{1}{x^{2}+3}$ times the derivative of $x^{2}+3$ , which is $2x$ .

Therefore, the first derivative of the function is $3(ln(x^{2}+3))^{2} * \frac{1}{x^{2}+3} * 2x$ .

To find the second derivative, apply the chain rule again to differentiate $3(ln(x^{2}+3))^{2}$ with respect to $x$ . The derivative of $3(ln(x^{2}+3))^{2}$ with respect to $u$ is $6u$ , where $u = ln(x^{2}+3)$ . Then, differentiate $ln(x^{2}+3)$ with respect to $x$ to get $\frac{1}{x^{2}+3} * 2x$ again.

Thus, the second derivative of the function is $6(ln(x^{2}+3)) * \frac{1}{x^{2}+3} * 2x$ .

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Answer from HIX Tutor

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