Answer:
The value of [tex]\frac{dx}{dt}[/tex] is [tex]\frac{160}{x^3}[/tex].
Step-by-step explanation:
The given equation is
[tex]x^4=8y^5-240[/tex]
We need to find the value of [tex]\frac{dx}{dt}[/tex].
Differentiate with respect to t.
[tex]4x^3\frac{dx}{dt}=8(5y^4)\frac{dy}{dt}-0[/tex] [tex][\because \frac{d}{dx}x^n=nx^{n-1},\frac{d}{dx}C=0][/tex]
[tex]4x^3\frac{dx}{dt}=40y^4\frac{dy}{dt}[/tex]
It is given that y=2 and dy/dt=1, substitute these values in the above equation.
[tex]4x^3\frac{dx}{dt}=40(2)^4(1)[/tex]
[tex]4x^3\frac{dx}{dt}=40(16)(1)[/tex]
[tex]4x^3\frac{dx}{dt}=640[/tex]
Divide both sides by 4x³.
[tex]\frac{dx}{dt}=\frac{640}{4x^3}[/tex]
[tex]\frac{dx}{dt}=\frac{160}{x^3}[/tex]
Therefore the value of [tex]\frac{dx}{dt}[/tex] is [tex]\frac{160}{x^3}[/tex].
