Derivatives and integrals help

[polulech]

derivative of integral over e^t to t^5 (sqrt(8+x^4)) dx

I know I need to use the chain rule and I can take the derivative of the integral without respect to e^t and t^5. If you know the answer, can you answer and tell me how to do it?! Calculus final on Monday...

 

[Ray Vickson]

derivative of integral over e^t to t^5 (sqrt(8+x^4)) dx

I know I need to use the chain rule and I can take the derivative of the integral without respect to e^t and t^5. If you know the answer, can you answer and tell me how to do it?! Calculus final on Monday...

Have you never seen the formula $$\frac{d}{dt}\int_{A(t)}^{B(t)} f(x,t) dx =<br /> \frac{dB(t)}{dt} \left. f(x,t)\right|_{x=B(t)} <br /> - \frac{dA(t)}{dt} \left. f(x,t)\right|_{x=A(t)} <br /> + \int_{A(t)}^{B(t)} \frac{\partial f(x,t)}{\partial t} \, dx ?$$

RGV

 

[polulech]

Umm no I have not :(
How do I put the pieces from my integral into that?

 

[SammyS]

derivative of integral over e^t to t^5 (sqrt(8+x^4)) dx

I know I need to use the chain rule and I can take the derivative of the integral without respect to e^t and t^5. If you know the answer, can you answer and tell me how to do it?! Calculus final on Monday...

Hi polulech. Welcome to PF.

Suppose F(x) is the anti-derivative of $\displaystyle \sqrt{8+x^4\,}\,.$

Then by the fundamental theorem of calculus, $\displaystyle \int_a^b\sqrt{8+x^4\,}\,dx=F(b)-F(a)\,.$

In the case of your integral you have: $\displaystyle \int_{e^t}^{\,t^5}\sqrt{8+x^4\,} \,dx=F(t^5)-F(e^t)\,.$

You know that $\displaystyle \frac{d}{dx}F(x)=\sqrt{8+x^4\,}\,.$ Combine this result with the chain rule to find the derivative of your integral.

 

[polulech]

how would I do that

 

[SammyS]

how would I do that

Can you find $\displaystyle \frac{d}{dt}F(t^5)\,,$ if you know that F'(x)=√(8 + x⁴) ?

 

[polulech]

would i plug t^5 into the √8+x^4 and then calculate the derivative?

 

[SammyS]

would i plug t^5 into the √8+x^4 and then calculate the derivative?

No.

F(x) is a function whose derivative is √(8+x⁴).

The chain rule says that $\displaystyle\frac{d}{dt}F(t^5)=F\,'(t^5)\cdot \frac{d}{dt}(t^5)\,.$

Furthermore, $\displaystyle F\,'(t^5)=\sqrt{8+t^{20}}$