contestada

Let F(x) = the integral from 0 to 2x of tan(t^2)dt . Use your calculator to find F"(1).
5.774
11.549
18.724
37.449

Respuesta :

LammettHash
By the fundamental theorem of calculus,

[tex]F(x)=\displaystyle\int_0^{2x}\tan(t^2)\,\mathrm dt[/tex]
[tex]\implies F'(x)=\dfrac{\mathrm d}{\mathrm dx}\displaystyle\int_0^{2x}\tan(t^2)\,\mathrm dt=\tan((2x)^2)\dfrac{\mathrm d}{\mathrm dx}[2x]=2\tan(4x^2)[/tex]

Then differentiating once more we find

[tex]F''(x)=2\sec^2(4x^2)\dfrac{\mathrm d}{\mathrm dx}[4x^2]=16x\sec^2(4x^2)[/tex]

and so [tex]F''(1)=16\sec^24\approx37.4488[/tex].

Let F(x) = the integral from 0 to 2x of tan(t^2)dt  then  F"(1) would be equal to 37.44.

What is chain rule of differentiation?

If a variable z depends on the variable y, which itself depends on the variable x, then we have:

[tex]\dfrac{dz}{dx} = \dfrac{dz}{dy} \times \dfrac{dy}{dx}[/tex]

(assuming all these derivatives exist).

By the fundamental theorem of calculus

[tex]f(x) = \int\limits^{2x}_0 tan(t^2)dt\\\\\\f'(x) = \dfrac{d}{dx} \int\limits^{2x}_0 tan(t^2)dt\\\\\\\\f'(x) = tan(2x^2)\dfrac{d}{dx} 2x\\\\=2 tan(4x^2)[/tex]

Then differentiating once more we find

[tex]f"(x) = 2sec^2(4x^2)\dfrac{d}{dx} 4x^2\\\\f"(x) = 16xsec^2(4x^2)\\\\f(1) = 16sec^4\\ = 37.44[/tex]

Learn more about chain rules here:

https://brainly.com/question/27462977

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