I'm having a little trouble with the following problem:
Calculate $F'(x)$:
$F(x)=\int_{1}^{x^{2}}(t-\sin^{2}t) dt$
It says we have to use substitution but I don't see why the answer can't just be:
$x-\sin^{2}x$
$\endgroup$ 15 Answers
$\begingroup$If you had $G(x)=\int_{1}^{x}(t-\sin^{2}t) dt$ (note that it's just $x$ at the upper limit of the integral), then $G'(x)=x-\sin^2x$. But, you have $F(x)=\int_{1}^{x^{2}}(t-\sin^{2}t) dt=G(x^2)$, so $F'(x)=G'(x^2)\cdot2x=(x^2-\sin^2x^2)2x$ (using the chain rule).
$\endgroup$ 1 $\begingroup$You could apply the following more general rule (Leibniz rule and chain rule) -- although Isaac's answer is more straightforward for this derivative: If [edited in response to a Didier Piau's comment to this answer of mine]
$$I(x)=J(u(x),v(x),x),\quad \text{with}\ J(a,b,x)=\int_a^bf(t,x)dt,$$
then, under suitable conditions, we have
$$I^{\prime }(x)=\displaystyle\int_{u(x)}^{v(x)}\dfrac{\partial f(t,x)}{\partial x}dt+f(v(x),x)v^{\prime }(x)-f(u(x),x)u^{\prime }(x).$$
In this case:
$$F(x)=I(x)=\displaystyle\int_{1}^{x^{2}}(t-\sin ^{2}t)dt,$$
and
$$u(x)=1,\quad v(x)=x^{2},\quad f(t,x)=(t-\sin ^{2}t),$$
$$\dfrac{\partial f(t,x)}{\partial x}=0,\quad v^{\prime }(x)=2x,\quad u^{\prime }(x)=0.$$
Therefore
$$F^{\prime }(x)=I^{\prime }(x)=(x^{2}-\sin ^{2}x^{2})2x=2x^{3}-2x\sin ^{2}x^{2}.$$
The conditions are: $u(x),v(x)$ are differentiable functions, $f(t,x)$ is a continuous function and $\partial f/\partial x$ exists and is continuous.
$\endgroup$ $\begingroup$Because the upper limit of the integral is x^2, not x.
$\endgroup$ $\begingroup$Well according to me the answer is $$F'(x) = \frac{d}{dx}(x^{2}) \cdot [x^{2}-\sin^{2}(x^{2})] - \frac{d}{dx}(1) \times \text{something} = 2x \cdot \Bigl[x^{2} -\sin^{2}(x^{2})\Bigr] - 0$$ would be the answer.
$\endgroup$ $\begingroup$$\displaystyle{% {{\rm d} \over {\rm d}x} = 2x\,{{\rm d} \over {\rm d}\left(x^{2}\right)} }$
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