# logarithmic differentiation

• Feb 20th 2009, 02:02 AM
katieeej
logarithmic differentiation
i'm so sorry for keep asking questions but im really confused for this one.
the problem is y=(e^(x-1)*sin^2*x)/(((x^2)+5)^2x)

the trig function (sin^2 + 5), i don't know how to get a derivative out of that.

help me plz!
• Feb 20th 2009, 02:10 AM
tah
Quote:

Originally Posted by katieeej
i'm so sorry for keep asking questions but im really confused for this one.
the problem is y=(e^(x-1)*sin^2*x)/(((x^2)+5)^2x)

the trig function (sin^2 + 5), i don't know how to get a derivative out of that.

help me plz!

Hi, may be there is a typo on the definition of the trigonometric function, is it sin^(x) + 5 or sin^2(x+5) or y is the variable or, ...May be I am missing the question
• Feb 20th 2009, 02:15 AM
katieeej
the equation is

e^(x-1)*sin^2 (x) / (x^2 + 5)^2x

its really complicated i'm sorry!
it's sin square of (x) and (x^2 +5) is x square and plus 5, separately.

thanks!
• Feb 20th 2009, 02:44 AM
synclastica_86
We have:
$y=\frac{e^{x-1}\sin^2{(x)}}{(x^2+5)^{2x}}={e^{x-1}\sin^2{(x)}}{(x^2+5)^{-2x}}$

To make things cleaner, we use the product rule and let:
$f(x)=e^{x-1}\sin^2(x)$
and
$g(x)=(x^2+5)^{-2x}=e^{\ln(x^2+5)^{-2x}}=e^{-2x\ln(x^2+5)}$

We now have:
$\frac{dy}{dx}=f'(x)g(x)+g'(x)f(x)$

Using the product:
$f'(x)=e^{x-1}\sin^2(x)+2\sin(x)\cos(x)e^{x-1}=e^{x-1}\sin(x)(2\cos(x)+\sin(x))$

Using the chain rule:
$g'(x)=e^{-2x\ln(x^2+5)}\frac{d}{dx}-2x\ln(x^2+5)$
$=-2(x^2+5)^{-2x}\frac{d}{dx}x\ln(x^2+5)$
$=-2(x^2+5)^{-2x}(x\frac{d}{dx}\ln(x^2+5)+\ln(x^2+5))$
$=-2(x^2+5)^{-2x}(\frac{x}{x^2+5}(2x)+\ln(x^2+5))$ here I use the property: $\frac{d}{du}\ln(u)=\frac{1}{u}$
$=-2(x^2+5)^{-2x}(\frac{2x^2}{x^2+5}+\ln(x^2+5))$

Now I'll leave it for you to plug in and simplify.
• Feb 20th 2009, 02:49 AM
tah
Quote:

Originally Posted by katieeej
the equation is

e^(x-1)*sin^2 (x) / (x^2 + 5)^2x

its really complicated i'm sorry!
it's sin square of (x) and (x^2 +5) is x square and plus 5, separately.

thanks!

the derivative looks a bit ugly but it follows from the formulas
$(uv)' = u'v + uv'\ and\ (\frac{u}{v})' = \frac{u'v-uv'}{v^2}$
which I am sure you have already used.

Try it out and compare your result to

$\frac{e^{x-1}sin(x) (x^3 sin(x) - 5 x^2sin(x) + 5 x sin(x) - 5 sin(x) + 2 x^3cos(x) + 10 x cos(x))}{x ^2(x ^2+ 5)^3 }$

which was done by a symbolic manipulation software (Yes I didn't do it :))
• Feb 20th 2009, 02:59 AM
Differentiating
Hello katieeej
Quote:

Originally Posted by katieeej
the equation is

e^(x-1)*sin^2 (x) / (x^2 + 5)^2x

its really complicated i'm sorry!
it's sin square of (x) and (x^2 +5) is x square and plus 5, separately.

thanks!

I'm sure you'll find it easier to begin by taking logs:

$ln(y) = (x-1) + 2ln(\sin x) - 2x \cdot ln(x^2+5)$

Then differentiate:

$\frac{1}{y}\frac{dy}{dx} = ...$ etc.

and finally multiply both sides by $y$ (= ...original function of $x$).