# Calculating a deriv using logarithmic differentiation

1. Feb 22, 2008

### danielle36

Hello,This is my first crack at using log differentiation, but I can't seem to get too far with it...

Use logarithmic differentiation to calculate the derivative for the following function:

$$y = \sqrt{x}e^{x^{2}} (x^{2} + 1)^{10}$$

$$lny = \frac{1}{2}lnx * x^{2}lne * 10ln(x^{2} + 1)$$

(I'm not sure what to do with the deriv of lne^{x^{2}})

$$lny = \frac{1}{2} * \frac{1}{x} * [deriv of lne^{x^{2}}] * 10 *\frac{2x}{x^{2} + 1}$$

$$lny = \frac{20x}{2x(x^{2} + 10} * [deriv of lne^{x^{2}}]$$

2. Feb 22, 2008

### VietDao29

I am assuming that you use '*' for multiplication.

You can notice that, taking derivative of summation is 'easier' than those of multiplication.

Say: (u + v + w)' = u' + v' + w' looks far easier compared with (uvw)' = u'vw + uv'w + uvw', right?

And using log function, one can change multiplication, into summation. Hence, making it more comfortable when dealing with a bunch of multiplication.

Nah, you are wrong from the first line :(, you should note that:

ln(xy) = ln(x) + ln(y), it's not ln(xy) = ln(x) * ln(y) as you've written.

I'll re-do it for you. Let's begin:

$$y = \sqrt{x} e ^ {x ^ 2} (x ^ 2 + 1) ^ {10}$$

Take logs of both sides yields:

$$\Rightarrow \ln( y ) = \ln \left( \sqrt{x} e ^ {x ^ 2} (x ^ 2 + 1) ^ {10} \right)$$

Use the formula: ln(xy) = ln(x) + ln(y), and ln(ab) = b ln(a) to expand all the terms on the RHS:

$$\Rightarrow \ln( y ) = \frac{1}{2} \ln (x) + x ^ 2 \ln (e) + 10 \ln (x ^ 2 + 1)$$

ln(e) = 1, so, let's simplify it a bit:

$$\Rightarrow \ln( y ) = \frac{1}{2} \ln (x) + x ^ 2 + 10 \ln (x ^ 2 + 1)$$

If the two functions (say, f(x), and g(x)) are the same (f(x) = g(x), for all x), then so are their derivatives (i.e, f'(x) = g'(x), for all x).

So, taking the derivatives of both sides with respect to x (not just one side as you did), we have:

$$\Rightarrow \frac{y'_x}{y} = \left( \frac{1}{2} \ln (x) + x ^ 2 + 10 \ln (x ^ 2 + 1) \right)'_x$$

Now, what you have to do is to take the derivatives of the RHS, then multiply y over, and arrive at the desired answer.

Hope that you can go from here. :)

--------------------------------

Btw, you should put a '\' in front of a function in LaTeX to make it looks un-italics.

Compare the two:

\ln x returns: $$\ln x$$

whereas ln x returns: $$ln x$$

The first one looks somewhat better, eh? :)

And to insert texts in LaTeX, we use the function \mbox{your text here}.

$$\mbox{LaTeX is wonderful :D}$$ is way better than $$LaTeX is wonderful :D$$. See? :)

Hint: Click on LaTeX images to see ther codes..

Last edited: Feb 23, 2008