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Assume that

x equals x left parenthesis t right parenthesisx=x(t)

and

y equals y left parenthesis t right parenthesisy=y(t).

Let

y equals x cubed plus 1y=x3+1

and

StartFraction dx Over dt EndFractiondxdt​=33

when

xequals=22.

Find

StartFraction dy Over dt EndFractiondydt

when

xequals=2

StartFraction dy Over dt EndFractiondydtequals=

Respuesta :

The value of [tex]\frac{dy}{dt}[/tex] is [tex]396[/tex]

Explanation:

It is given that [tex]x=x(t)[/tex] and [tex]y=y(t)[/tex]

Also, given that [tex]y=x^3+1[/tex] and [tex]\frac{dx}{dt} =33[/tex] when [tex]x=22[/tex]

Now, we need to determine the value of [tex]\frac{dy}{dt}[/tex] when [tex]x=2[/tex]

To determine the value of [tex]\frac{dy}{dt}[/tex] when [tex]x=2[/tex], we need to differentiate the function [tex]y=x^3+1[/tex] with respect to [tex]\frac{dy}{dt}[/tex]

Thus, we have,

[tex]\frac{dy}{dt}=3x^2(\frac{dx}{dt} )+0[/tex]

Simplifying, we get,

[tex]\frac{dy}{dt}=3x^2(\frac{dx}{dt} )[/tex]

Substituting [tex]x=2[/tex] and [tex]\frac{dx}{dt} =33[/tex] , we get,

[tex]\frac{dy}{dt}=3(2)^2(33 )[/tex]

Squaring the term, we have,

[tex]\frac{dy}{dt}=3(4)(33 )[/tex]

Multiplying the terms, we get,

[tex]\frac{dy}{dt}=396[/tex]

Thus, the value of [tex]\frac{dy}{dt}[/tex] is [tex]396[/tex]

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