Derivative of Integral

I'm having a little problem with the adhering to trouble:

Calculate $F'(x)$:

$F(x)=\int_{1}^{x^{2}}(t-\sin^{2}t) dt$

It claims we need to make use of replacement yet I do not see why the solution can not simply be:

$x-\sin^{2}x$

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2019-05-18 22:12:14
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Answers: 4

If you had $G(x)=\int_{1}^{x}(t-\sin^{2}t) dt$ (note that it is simply $x$ at the ceiling of the indispensable), after that $G'(x)=x-\sin^2x$. Yet, you have $F(x)=\int_{1}^{x^{2}}(t-\sin^{2}t) dt=G(x^2)$, so $F'(x)=G'(x^2)\cdot2x=(x^2-\sin^2x^2)2x$ (making use of the chain regulation).

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2019-05-21 07:26:40
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Because the ceiling of the indispensable is x ^ 2, not x.

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2019-05-21 07:25:47
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Well according to me the solution 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 certainly be the solution.

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2019-05-21 06:28:19
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You can use the adhering to extra basic regulation (Leibniz rule and also chain rule) - - although Isaac is solution is extra uncomplicated for this by-product: If [ modified in feedback to a Didier Piau is comment to this solution of mine ]

$$I(x)=J(u(x),v(x),x),\quad \text{with}\ J(a,b,x)=\int_a^bf(t,x)dt,$$

after that, under ideal problems, 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 instance:

$$F(x)=I(x)=\displaystyle\int_{1}^{x^{2}}(t-\sin ^{2}t)dt,$$

and also

$$u(x)=1,\quad v(x)=x^{2},\quad f(t,x)=(t-\sin ^{2}t),$$

$$\dfrac{\partial f(t,x)}{\partial x}=0,\quad v^{\prime }(x)=2x,\quad u^{\prime }(x)=0.$$

Therefore

$$F^{\prime }(x)=I^{\prime }(x)=(x^{2}-\sin ^{2}x^{2})2x=2x^{3}-2x\sin ^{2}x^{2}.$$


The problems are: $u(x),v(x)$ are differentiable features, $f(t,x)$ is a. continual function and also $\partial f/\partial x$ exists and also is continual.

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2019-05-21 05:54:11
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