Question

F(3) = 8; f^ prime prime (3)=-4; g(3)=2,g^ prime (3)=-6 , find F(3) if F(x) = root(4, f(x) * g(x))

Answer 1

Given:

`f(3)=8,f^(\prime)(3)=-4,g(3)=2,\text{ and }g^(\prime)(3)=-6`

Required:

`We\text{ need to find }F^(\prime)(3)\text{ if }F(x)=\sqrt[4]{f(x)g(x)}.`

Step-by-step explanation:

Given equation is

`F(x)=\sqrt[4]{f(x)g(x)}.`
`F(x)=(f(x)g(x))^{(1)/(4)}`
`F(x)=f(x)^{(1)/(4)}g(x)^{(1)/(4)}`

Differentiate the given equation for x.

`Use\text{ }(uv)^(\prime)=uv^(\prime)+vu^(\prime).\text{ Here u=}\sqrt[4]{f(x)}\text{ and v=}\sqrt[4]{g(x)}.`

`F^(\prime)(x)=f(x)^{(1)/(4)}((1)/(4)g(x)^{(1)/(4)-1})g^(\prime)(x)+g(x)^{(1)/(4)}((1)/(4)f(x)^{(1)/(4)-1})f^(\prime)(x)`
`=(1)/(4)f(x)^{(1)/(4)}g(x)^{(1)/(4)-(1*4)/(4)}g^(\prime)(x)+(1)/(4)g(x)^{(1)/(4)}f(x)^{(1)/(1)-(1*4)/(4)}f^(\prime)(x)`
`=(1)/(4)f(x)^{(1)/(4)}g(x)^{(1-4)/(4)}g^(\prime)(x)+(1)/(4)g(x)^{(1)/(4)}f(x)^{(1-4)/(4)}f^(\prime)(x)`
`F^(\prime)(x)=(1)/(4)f(x)^{(1)/(4)}g(x)^{(-3)/(4)}g^(\prime)(x)+(1)/(4)g(x)^{(1)/(4)}f(x)^{(-3)/(4)}f^(\prime)(x)`

Replace x=3 in the equation.

`F^(\prime)(3)=(1)/(4)f(3)^{(1)/(4)}g(3)^{(-3)/(4)}g^(\prime)(3)+(1)/(4)g(3)^{(1)/(4)}f(3)^{(-3)/(4)}f^(\prime)(3)`
`Substitute\text{ }f(3)=8,f^(\prime)(3)=-4,g(3)=2,\text{ and }g^(\prime)(3)=-6\text{ in the equation.}`
`F^(\prime)(3)=(1)/(4)(8)^{(1)/(4)}(2)^{(-3)/(4)}(-6)+(1)/(4)(2)^{(1)/(4)}(8)^{(-3)/(4)}(-4)`
`F^(\prime)(3)=(-6)/(4)(8)^{(1)/(4)}(2^3)^{(-1)/(4)}+(-4)/(4)(2)^{(1)/(4)}(8^3)^{(-1)/(4)}`
`F^(\prime)(3)=(-3)/(2)(8)^{(1)/(4)}(8)^{(-1)/(4)}-(2)^{(1)/(4)}(8^3)^{(-1)/(4)}`
`F^(\prime)(3)=(-3)/(2)\frac{\sqrt[4]{8}}{\sqrt[4]{8}}-\frac{\sqrt[4]{2}}{\sqrt[4]{8^3}}`
`F^(\prime)(3)=(-3)/(2)-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^9}}`
`F^(\prime)(3)=(-3)/(2)-\frac{\sqrt[4]{2}}{\sqrt[4]{(2)^4(2)^4}(2)}`
`F^(\prime)(3)=(-3)/(2)-\frac{\sqrt[4]{2}}{4\sqrt[4]{}(2)}`
`F^(\prime)(3)=(-3)/(2)-(1)/(4)`
`F^(\prime)(3)=(-3*2)/(2*2)-(1)/(4)`
`F^(\prime)(3)=(-6-1)/(4)`
`F^(\prime)(3)=(-7)/(4)`

Final answer:

`F^(\prime)(3)=(-7)/(4)`