Use part 1 of the fundamental theorem of calculus to find the derivative of the function. y = 5 u3 1 + u2 du 4 − 3x

Using part 1 of the fundamental theorem of calculus to find the derivative of the function. The derivative of the given function is:

[tex]\mathbf{\dfrac{dy}{dx} =3\Big (\dfrac{(4-3x)^3}{1+(4-3x)^2}\ \Big) \ }[/tex]

Consider the given function:

[tex]\mathbf{y = \int^5_{4-3x} \ \dfrac{u^3}{1+u^2}\ du}[/tex]

The objective is to find  [tex]\mathbf{\dfrac{dy}{dx}}[/tex]  by using the fundamental theorem of calculus.

Using chain rule:

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

[tex]\mathbf{ =\dfrac{d}{dv}\Big (\int^5_{4-3x} \dfrac{u^3}{1+u^2}\ du\Big) \dfrac{dv}{dx}}}[/tex]

[tex]\mathbf{ =\dfrac{d}{dv}\Big (\int^5_{v} \dfrac{u^3}{1+u^2}\ du\Big) \dfrac{dv}{dx} \ \ \ \ \ since \ v \ = 4 - 3x} }[/tex]

[tex]\mathbf{ =-\dfrac{d}{dv}\Big (\int^v_{5} \dfrac{u^3}{1+u^2}\ du\Big) \dfrac{dv}{dx} }[/tex]

[tex]\mathbf{ =-\dfrac{d}{dv}\Big (\int^v_{5} \dfrac{u^3}{1+u^2}\ du\Big) \ (-3)}[/tex]

[tex]\mathbf{ =3\dfrac{d}{dv}\Big (\int^v_{5} \dfrac{u^3}{1+u^2}\ du\Big) \ }[/tex]

From the fundamental theorem of calculus;

 [tex]\mathbf{\dfrac{d}{dx} \Big( \int^x_1 \ g(t) dt \Big) = g(x)}[/tex]

∴

[tex]\mathbf{ =3\dfrac{d}{dv}\Big (\int^v_{5} \dfrac{u^3}{1+u^2}\ du\Big) \ }[/tex] will be:

[tex]\mathbf{ =3\times \Big (\dfrac{(4-3x)^3}{1+(4-3x)^2}\ \Big) \ }[/tex]

∴

[tex]\mathbf{\dfrac{dy}{dx} =3\Big (\dfrac{(4-3x)^3}{1+(4-3x)^2}\ \Big) \ }[/tex]

Therefore, we can conclude that the derivative of [tex]\mathbf{y = \int^5_{4-3x} \ \dfrac{u^3}{1+u^2}\ du}[/tex]

using the fundamental theorem of calculus is   [tex]\mathbf{\dfrac{dy}{dx} =3\Big (\dfrac{(4-3x)^3}{1+(4-3x)^2}\ \Big) \ }[/tex]

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