Question

Find the derivative of the function using the appropriate form the fundamental theorem of calculus.

Answer 1

\[F(x) = \int\limits_{\sqrt{x}}^{1}\frac{ s^2 }{ 2 +5s^4 }ds\]

Answer 2

so you can do a trick so it is \[-\int\limits_{1}^{\sqrt{x}}\frac{ s^2 }{ 2+5s^4 }ds\]

Answer 3

then wouldn't you just plug in sqrt(x) into the function and get \[\frac{ -\sqrt{x}^2 }{ 2 + 5x^4 }dx\]

Answer 4

but that is coming up as incorrect

Answer 5

Well, you're on the right path, that trick is a good one, but remember you need to do the chain rule and take the derivative of the square root as well.

Answer 6

\[F(x)=\int\limits_{a(x)}^{b(x)}g(s) ds \\ F(x)=G(s)|_{a(x)}^{b(x)} \text{ note : where } G'=g \\ F(x)=G(b(x))-G(a(x)) \\ \text{ now we get \to differentiate both sides } \\ F'(x)=(G(b(x))-G(a(x)))' \\ \text{ by difference rule we have } \\ F'(x)=(G(b(x))'-(G(a(x))' \ \text{ now you apply chain rule }\]

Answer 7

So to fix it, I noticed you also put a dx there, but that is part of the integral so when you take the derivative it will go away. Writing in the chain rule, it will look like this, and it's almost done. \[\frac{ -\sqrt{x}^2 }{ 2 + 5x^4 } ( \sqrt{x})'\]

Answer 8

okay so I need to add a \[- \frac{ 1^2 }{ 2+ 5 (1)^4 } (1)'\] ?

Answer 9

you can but (1)'=0 so no need really