need help with the following calculus function

[integragirl]

I'm supposed to find dy/dx y=sqrt {(x+1)^10}/(2x+5)^5 the whole equation is under one squareroot sign. Any help would be appreciated

 

[stapel]

I'm supposed to find dy/dx y=sqrt {(x+1)^10}/(2x+5)^5

Do you mean that you are supposed to find dy/dx given that y is the posted expression...? (I'm not familiar with "dy/dxy" or "(dy/dx)y", is why I ask.)

And is y equal to the following?

. . . . .$\displaystyle \large{y\,= \,\sqrt{\frac{(x\,+\,1)^{10}}{(2x\,+\,5)^5}}}$

What rules are you familiar with? Have you learned the Quotient Rule? The Power Rule (for differentiating square roots, being the one-half power)? The Chain Rule?

Thank you.

Eliz.

 

[integragirl]

yes, you have the equation posted correctly. We are covering basic integrals in class and I believe that this equation is somehow related. We have covered the power rule and the quotient rule but I don't really understand the chain rule.

 

[stapel]

We are covering basic integrals in class...

So... are you integrating or differentiating?

For an explanation of the Chain Rule, try these:

. . . . .Paul's Online Notes: The Chain Rule

. . . . .Ask Dr. Math: Trig and the Chain Rule

. . . . .old newsgroup discussion of Chain Rule

Thank you.

Eliz.

 

[integragirl]

integrating

 

[skeeter]

$\displaystyle \large{y\,= \,\sqrt{\frac{(x\,+\,1)^{10}}{(2x\,+\,5)^5}}}$

$\displaystyle \large{y\,=\,\frac{(x\,+\,1)^5}{(2x\,+\,5)^{\frac{5}{2}}}}$

$\displaystyle \large{\frac{dy}{dx}\,=\,\frac{(2x+5)^{\frac{5}{2}}5(x+1)^4\,-\,(x+1)^55(2x+5)^{\frac{3}{2}}}{(2x+5)^5}}$

\(\displaystyle \large{\frac{dy}{dx}\.=\,\frac{5(2x+5)^{\frac{3}{2}}(x+1)^4[(2x+5)\,-\,(x+1)]}{(2x+5)^5}}\)

$\displaystyle \large{\frac{dy}{dx}\,=\,\frac{5(x+1)^4(x+4)}{(2x+5)^{\frac{7}{2}}}}$

$\displaystyle \large{\frac{dy}{dx}\,=\,\frac{5(x+1)^4(x+4)}{\sqrt{(2x+5)^7}}}$

 

[stapel]

integrating

So you're actually needing to find the indefinite integral of the square-root expression, rather than dy/dx?

Eliz.