What is the derivative of f(x)=(2x+5)/(x+2)*2x-5 ?

The derivative of f(x)=(2x+5)/(x+2)*2x-5 is (2(2x^2+8x+10))/((x+2)^2)

What is the first derivative of f(x)=(2x+5)/(x+2)*2x-5 ?

The first derivative of f(x)=(2x+5)/(x+2)*2x-5 is (2(2x^2+8x+10))/((x+2)^2)

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derivative of

f (x) = \frac{2 x + 5}{x + 2} \cdot 2 x - 5

\frac{2 ( 2 x ^{2} + 8 x + 10 )}{( x + 2 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{2 x + 5}{x + 2} \cdot 2 x - 5)

Simplify

\frac{2 x + 5}{x + 2} \cdot 2 x - 5 : \frac{2 x ( 2 x + 5 )}{x + 2} - 5

= \frac{d}{d x} (\frac{2 x ( 2 x + 5 )}{x + 2} - 5)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (\frac{2 x ( 2 x + 5 )}{x + 2}) - \frac{d}{d x} (5)

\frac{d}{d x} (\frac{2 x ( 2 x + 5 )}{x + 2}) = \frac{2 ( 2 x ^{2} + 8 x + 10 )}{( x + 2 ) ^{2}}

\frac{d}{d x} (5) = 0

= \frac{2 ( 2 x ^{2} + 8 x + 10 )}{( x + 2 ) ^{2}} - 0

Simplify

= \frac{2 ( 2 x ^{2} + 8 x + 10 )}{( x + 2 ) ^{2}}

\frac{d y}{d x} = x y + y^{2}

\int \frac{12 x ^{2}}{4 x ^{3} - 6} d x

y^{^{'}} = \frac{( 3 x sec ( \frac{y}{x} ) + y )}{x}

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

\frac{\partial}{\partial y} (x y^{4})

What is the derivative of f(x)=(2x+5)/(x+2)*2x-5 ?
The derivative of f(x)=(2x+5)/(x+2)*2x-5 is (2(2x^2+8x+10))/((x+2)^2)
What is the first derivative of f(x)=(2x+5)/(x+2)*2x-5 ?
The first derivative of f(x)=(2x+5)/(x+2)*2x-5 is (2(2x^2+8x+10))/((x+2)^2)