- #1
Eclair_de_XII
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- 91
Homework Statement
"Under mild continuity restrictions, it is true that if ##F(x)=\int_a^b g(t,x)dt##,
then ##F'(x)=\int_a^b g_x(t,x)dt##.
Using this fact and the Chain Rule, we can find the derivative of
##F(x)=\int_{a}^{f(x)} g(t,x)dt##
by letting
##G(u,x)=\int_a^u g(t,x)dt##,
where ##u=f(x)##."
Given:
##F(x)=\int_{0}^{x^2} \sqrt{t^4+x^3}dt##,
find ##\frac{dF}{dx}##.
Homework Equations
##\frac{dF}{dx}=\frac{dF}{dt}\frac{dt}{dx}+\frac{dF}{du}\frac{du}{dx}##
The Attempt at a Solution
##\frac{dF}{du}=\frac{d}{du} \int_{0}^{x^2}\sqrt{t^4+x^3}dt=\sqrt{(x^2)^4+x^3}=\sqrt{x^8+x^3}##
##\frac{du}{dx}=\frac{d}{dx}(x^2)=2x## by letting ##u=x^2##
##\frac{dF}{dt}=F'(t)=\int_{0}^{x^2} g_x(t^4+x^3)dt=\int_{0}^{x^2} \frac{3x^2}{2\sqrt{t^4+x^3}}##
I know that: ##\frac{dF}{dx}=2x\sqrt{x^8+x^3}+\int_{0}^{x^2} \frac{3x^2}{2\sqrt{t^4+x^3}}\frac{dt}{dx}##
But, I'm at a loss at how to calculate: ##\frac{dt}{dx}##, because I don't know how to define ##t##, so I don't know how to differentiate it with respect to ##x##.