# Thread: A couple of logarithmic differentiation questions

1. ## A couple of logarithmic differentiation questions

I'm not sure how to derive these:

y = log(2)x^(1/3) (2)=base

and

y = √x(e^3x²)

2. Originally Posted by b521
I'm not sure how to derive these:

y = log(2)x^(1/3) (2)=base

and

y = √x(e^3x²)

Rewrite the given equation:

$y = \log_2(x^{\frac13})~\implies~y=\frac13 \cdot \frac1{\ln(2)} \cdot \ln(x)$

You only have to know how $\ln(x)$ is differentiated, all other factors are constant.

Do the second question similarly.

3. Only ln(x) is differentiated? So it would be 1/3xln2 ?

4. Well, with parentheses to make it clearer: 1/(3(ln 2)x). Yes, that is correct.

In general, the derivative of $e^x$ is $e^x$ while the derivative of $a^x$ with a any positive number, is $a^x ln(a)$. The derivative of $ln(x)$ is $\frac{1}{x}$ while the derivative of $log_a(x)$ with a any positive number is $\frac{1}{x ln(a)}$.

5. So for the second equation, would the answer use fg'+gf'?

6ln(3x²)x^(1/2) + 1/(2x^(1/2))e^(3x²)
(6√x)ln(3x²) + e^(3x²)/(2√x)

6. Originally Posted by b521
So for the second equation, would the answer use fg'+gf'?

6ln(3x²)x^(1/2) + 1/(2x^(1/2))e^(3x²)
(6√x)ln(3x²) + e^(3x²)/(2√x)

1. I assume that the given function reads:

$y = \sqrt{x} \cdot e^{3x^2}$

2. Re-write so that you have only one power to the base e at the RHS:

$y = e^{3x^2+\frac12 \ln(x)}$

3. Now differentiate using the chain rule:

$y'= e^{3x^2+\frac12 \ln(x)} \cdot \left(6x+\frac1{2x}\right)$

Simplify!

4. If you want to use the product rule and the chain rule you'll get:

$y = \sqrt{x} \cdot e^{3x^2}~\implies~y'=e^{3x^2} \cdot \frac1{2\sqrt{x}} + \sqrt{x} \cdot e^{3x^2} \cdot 6x$

Simplify!