What is the derivative of y=((f(x))/(g(x)))^2 ?

The derivative of y=((f(x))/(g(x)))^2 is (2f(x)(\frac{df)/(dx)g(x)-(dg)/(dx)f(x))}{g(x)^3}

What is the first derivative of y=((f(x))/(g(x)))^2 ?

The first derivative of y=((f(x))/(g(x)))^2 is (2f(x)(\frac{df)/(dx)g(x)-(dg)/(dx)f(x))}{g(x)^3}

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derivative of

y = (\frac{f ( x )}{g ( x )})^{2}

\frac{2 f ( x ) ( \frac{df}{d x} g ( x ) - \frac{d g}{d x} f ( x ) )}{g ( x ) ^{3}}

Solution steps

\frac{d}{d x} ((\frac{f ( x )}{g ( x )})^{2})

Apply the chain rule:

2 \cdot \frac{f ( x )}{g ( x )} \frac{d}{d x} (\frac{f ( x )}{g ( x )})

= 2 \cdot \frac{f ( x )}{g ( x )} \frac{d}{d x} (\frac{f ( x )}{g ( x )})

\frac{d}{d x} (\frac{f ( x )}{g ( x )}) = \frac{\frac{df}{d x} g ( x ) - \frac{d g}{d x} f ( x )}{g ( x ) ^{2}}

= 2 \cdot \frac{f ( x )}{g ( x )} \cdot \frac{\frac{df}{d x} g ( x ) - \frac{d g}{d x} f ( x )}{g ( x ) ^{2}}

= 2 \cdot \frac{f ( x )}{g ( x )} \cdot \frac{\frac{df}{d x} g ( x ) - \frac{d g}{d x} f ( x )}{g ( x ) ^{2}}

Simplify

2 \cdot \frac{f ( x )}{g ( x )} \cdot \frac{\frac{df}{d x} g ( x ) - \frac{d g}{d x} f ( x )}{g ( x ) ^{2}} : \frac{2 f ( x ) ( \frac{df}{d x} g ( x ) - \frac{d g}{d x} f ( x ) )}{g ( x ) ^{3}}

= \frac{2 f ( x ) ( \frac{df}{d x} g ( x ) - \frac{d g}{d x} f ( x ) )}{g ( x ) ^{3}}

\int x^{4} e^{2 x} d x

\int \frac{2 x + 1}{( 3 x - 1 ) ( x ^{2} + 2 x + 2 )} d x

\frac{\partial}{\partial x} (4 - 2 \cdot (3 x - y)^{2})

y = - (- 6 x + 6)^{\frac{1}{2}}, x = - 5

\frac{d}{d v} (a e^{v} + \frac{b}{v} + \frac{c}{v ^{2}})

What is the derivative of y=((f(x))/(g(x)))^2 ?
The derivative of y=((f(x))/(g(x)))^2 is (2f(x)(\frac{df)/(dx)g(x)-(dg)/(dx)f(x))}{g(x)^3}
What is the first derivative of y=((f(x))/(g(x)))^2 ?
The first derivative of y=((f(x))/(g(x)))^2 is (2f(x)(\frac{df)/(dx)g(x)-(dg)/(dx)f(x))}{g(x)^3}