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!
We have:
$\displaystyle y=\frac{e^{x-1}\sin^2{(x)}}{(x^2+5)^{2x}}={e^{x-1}\sin^2{(x)}}{(x^2+5)^{-2x}}$
To make things cleaner, we use the product rule and let:
$\displaystyle f(x)=e^{x-1}\sin^2(x)$
and
$\displaystyle g(x)=(x^2+5)^{-2x}=e^{\ln(x^2+5)^{-2x}}=e^{-2x\ln(x^2+5)}$
We now have:
$\displaystyle \frac{dy}{dx}=f'(x)g(x)+g'(x)f(x)$
Using the product:
$\displaystyle f'(x)=e^{x-1}\sin^2(x)+2\sin(x)\cos(x)e^{x-1}=e^{x-1}\sin(x)(2\cos(x)+\sin(x))$
Using the chain rule:
$\displaystyle g'(x)=e^{-2x\ln(x^2+5)}\frac{d}{dx}-2x\ln(x^2+5)$
$\displaystyle =-2(x^2+5)^{-2x}\frac{d}{dx}x\ln(x^2+5)$
$\displaystyle =-2(x^2+5)^{-2x}(x\frac{d}{dx}\ln(x^2+5)+\ln(x^2+5))$
$\displaystyle =-2(x^2+5)^{-2x}(\frac{x}{x^2+5}(2x)+\ln(x^2+5))$ here I use the property: $\displaystyle \frac{d}{du}\ln(u)=\frac{1}{u}$
$\displaystyle =-2(x^2+5)^{-2x}(\frac{2x^2}{x^2+5}+\ln(x^2+5))$
Now I'll leave it for you to plug in and simplify.
the derivative looks a bit ugly but it follows from the formulas
$\displaystyle (uv)' = u'v + uv'\ and\ (\frac{u}{v})' = \frac{u'v-uv'}{v^2}$
which I am sure you have already used.
Try it out and compare your result to
$\displaystyle \frac{e^{x-1}sin(x) (x^3 sin(x) - 5 x^2sin(x) + 5 x sin(x) - 5 sin(x) + 2 x^3cos(x) + 10 x cos(x))}{x ^2(x ^2+ 5)^3 }$
which was done by a symbolic manipulation software (Yes I didn't do it )
Hello katieeejI'm sure you'll find it easier to begin by taking logs:
$\displaystyle ln(y) = (x-1) + 2ln(\sin x) - 2x \cdot ln(x^2+5)$
Then differentiate:
$\displaystyle \frac{1}{y}\frac{dy}{dx} = ...$ etc.
and finally multiply both sides by $\displaystyle y$ (= ...original function of $\displaystyle x$).
Grandad