Fundamental theorem calculus with upper limit function of both x and y

If $y$ is a differentiable function of $x$ such that $(1 + x^3)y^2 + 4 \int_{2x}^{xy}(5x^2+t^2 )^{0.5} dt = 112$, find the value of $\frac{dy}{dx}$ at the point where $y = 2$. The solution is $\frac{-27}{\;55}$.

There is a lot going on and the parts giving trouble are the upper limit of integration equal to $xy$, and no value for $x$ when $y = 2$.

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If $y=2$, the integral is 0 and you can solve for $x$. For differentiating the integral, see here. – David Mitra Mar 31 '12 at 18:32

Your equation is $$\tag{1} (1 + x^3)y^2 + 4 \int_{2x}^{xy}(5x^2+t^2 )^{0.5} dt = 112.$$ Note that if $y=2$, then the integral in $(1)$ is zero since the lower and upper limits of integration are the same. You can then solve for $x$ and discover that it is 3 when $y=2$.

Of course, you'll want to implicitly differentiate equation $(1)$. But there is a subtlety here when differentiating the integral. The integrand in the term $$\Phi(x)=\int_{2x}^{xy} (5x^2+t^2)^{1/2}\,dt$$ is a function of $x$ and $t$; you cannot use the Fundamental Theorem of Calculus (which requires that the integrand is a function of $t$ only) directly to find its derivative. In particular, even if you split the integral into two parts $$\int_{2x}^{xy} (5x^2+t^2)^{1/2}\,dt= \int_{2x}^{0} (5x^2+t^2)^{1/2}\,dt+ \int_{0}^{xy} (5x^2+t^2)^{1/2}\,dt,$$ you cannot say, for example, that $${d\over dx} \int_{0}^{xy} (5x^2+t^2)^{1/2}\,dt= (5x^2+(xy)^2)^{1/2}\cdot {d\over dx}(xy).$$

However, to find the derivative of $\Phi$, you can use the technique of differentiation under the integral sign. Using this rule gives: \eqalign{ {d\over dx} \Phi(x)&= {d\over dx} \int_{2x}^{xy} (5x^2+t^2)^{1/2}\,dt\cr&= (5x^2+(xy)^2)^{1/2} (y+xy')-2(5x^2+4x^2)^{1/2}\, +\int_{2x}^{xy}{\partial\over\partial x} (5x^2+t^2)^{1/2}\,dt } (note the rule gives what you would obtain if you just applied FTOC plus an integral term).

After implicitly differentiating both sides of $(1)$, you should wind up with

$$\tag{2}3x^2y^2+(1+x^3)2y\cdot y'+4\textstyle {d\over dx} \Phi(x) .$$ When you evaluate $(2)$ at $y=2$, $x=3$, the integral term is zero, and you can then solve for $y'|_{y=2}$; thus, obtaining the posted solution.

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Thanks. I split the integral into two parts. First a negative version with limits a to 2x and a similar, but positive version with limits a to xy. This because I did not understand why you added the integral in your line (2) implicit differentiation. – Bob Apr 1 '12 at 11:32
@Bob It's required from the differentiation rule I linked. Note the integrand is a function of $x$ and $t$. – David Mitra Apr 1 '12 at 12:29