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About LinkBacksdifferentiate the first function
integrate the second one
how's this? can you help me please?
Well, the first one's simple enough. You can't integrate as long as you have a trig function in the exponent, so, do you remember how to get it down?
Just take the natural log of both sides.
$\displaystyle lny=ln2(sec 3x)(cosx+4x^3)$
Then take the derivative:
(1/y)(dy/dx)=[1/2(sec 3x)(cosx+4x^3)][6(sec 3x)(tan 3x)(cosx+4x^3)+2(sec 3x)(12x^2-sinx)]
But then you need to get rid of the $\displaystyle y$, right? So substitute the original equation back in by multiplying both sides by it:
(dy/dx)=[(cosx+4x^3)^{2sec3x}][1/2(sec 3x)(cosx+4x^3)][6(sec 3x)(tan 3x)(cosx+4x^3)+2(sec 3x)(12x^2-sinx)]
Please keep in mind I'm still getting the hang of this math-on-the-web notation, so I may have made a careless mistake resulting in a wrong final answer, but the principle is the same. Take the nat. log of both sides, then the derivatives. You'll be left with a $\displaystyle 1/y$ on the left side, which means you'll need to substitute back in the original equation when you're done, and multiply both sides by it. To take the derivative of the right side, use the chain rule (d/dx of ln u = u'/u), then the product rule (d/dx of u*v = u'v+uv'). Time-consuming, but simple.
I could probably do the second one, but I don't have time right now, sorry.
Hello, TrueToHeart!
Here's the second one . . .
We have: .$\displaystyle \int(2^{\ln x} -5)^{-10}\left(\frac{1}{x}\!\cdot\!2^{\ln x}\!\cdot\!dx\right)$$\displaystyle \int\frac{2^{\ln x}}{x(2^{\ln x-5})^{10}}\,dx$
Let $\displaystyle u \:=\:2^{\ln x} - 5\quad\Rightarrow\quad du \:=\:\frac{1}{x}\!\cdot\!2^{\ln x}\!\cdot\!\ln2\!\cdot\!dx \quad\Rightarrow\quad\frac{1}{x}\!\cdot\!2^{\ln x}\!\cdot\!dx \:=\:\frac{du}{\ln 2}$
Substitute: .$\displaystyle \int u^{-10}\!\cdot\!\frac{du}{\ln 2} \;\;=\;\;\frac{1}{\ln2}\int u^{-10}\,du\;\;=\;\;-\frac{1}{9}\,u^{-9} + C$
Back-substitute: . $\displaystyle -\frac{1}{9(2^{\ln x}- 5)^9} +C$
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