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

The derivative of f(x)=sqrt((2x^2-x)/(x+5)) is (2x^2+20x-5)/(2sqrt(2x^2-x)(x+5)^{3/2)}

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

The first derivative of f(x)=sqrt((2x^2-x)/(x+5)) is (2x^2+20x-5)/(2sqrt(2x^2-x)(x+5)^{3/2)}

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derivative of

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

\frac{2 x ^{2} + 20 x - 5}{2 2 x ^{2} - x ( x + 5 ) ^{\frac{3}{2}}}

Solution steps

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

Apply the chain rule:

\frac{1}{2 \frac{2 x ^{2} - x}{x + 5}} \frac{d}{d x} (\frac{2 x ^{2} - x}{x + 5})

= \frac{1}{2 \frac{2 x ^{2} - x}{x + 5}} \frac{d}{d x} (\frac{2 x ^{2} - x}{x + 5})

\frac{d}{d x} (\frac{2 x ^{2} - x}{x + 5}) = \frac{2 x ^{2} + 20 x - 5}{( x + 5 ) ^{2}}

= \frac{1}{2 \frac{2 x ^{2} - x}{x + 5}} \cdot \frac{2 x ^{2} + 20 x - 5}{( x + 5 ) ^{2}}

Simplify

\frac{1}{2 \frac{2 x ^{2} - x}{x + 5}} \cdot \frac{2 x ^{2} + 20 x - 5}{( x + 5 ) ^{2}} : \frac{2 x ^{2} + 20 x - 5}{2 2 x ^{2} - x ( x + 5 ) ^{\frac{3}{2}}}

= \frac{2 x ^{2} + 20 x - 5}{2 2 x ^{2} - x ( x + 5 ) ^{\frac{3}{2}}}

\int 3 arctan (x) d x

5 x^{2} + \frac{1}{x ^{2}}

\frac{x + 1}{x ^{3} + x - 3}

x \to 1 lim (x ln (x))

\int \frac{1}{x + 2} d x

What is the derivative of f(x)=sqrt((2x^2-x)/(x+5)) ?
The derivative of f(x)=sqrt((2x^2-x)/(x+5)) is (2x^2+20x-5)/(2sqrt(2x^2-x)(x+5)^{3/2)}
What is the first derivative of f(x)=sqrt((2x^2-x)/(x+5)) ?
The first derivative of f(x)=sqrt((2x^2-x)/(x+5)) is (2x^2+20x-5)/(2sqrt(2x^2-x)(x+5)^{3/2)}