What is the derivative of y=(1+xf(x))/(sqrt(x)) ?

The derivative of y=(1+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-1}{2xsqrt(x)}

What is the first derivative of y=(1+xf(x))/(sqrt(x)) ?

The first derivative of y=(1+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-1}{2xsqrt(x)}

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derivative of

y = \frac{1 + x f ( x )}{x}

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

Solution steps

\frac{d}{d x} (\frac{1 + x f ( x )}{x})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

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

\frac{d}{d x} (1 + x f (x)) = x \frac{df}{d x} + f (x)

\frac{d}{d x} (x) = \frac{1}{2 x}

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

Simplify

\frac{( x \frac{df}{d x} + f ( x ) ) x - \frac{1}{2 x} ( 1 + x f ( x ) )}{( x ) ^{2}} : \frac{2 x ^{2} \frac{df}{d x} + x f ( x ) - 1}{2 x x}

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

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

\int \frac{8 + 7 x}{1 + x ^{2}} d x

f (x) = x^{2} - 2, at x = 0

\int \frac{2 x ^{3} - 4 x ^{2} + 18 x}{x ^{2} + 9} d x

\int x x x + 7 d x

What is the derivative of y=(1+xf(x))/(sqrt(x)) ?
The derivative of y=(1+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-1}{2xsqrt(x)}
What is the first derivative of y=(1+xf(x))/(sqrt(x)) ?
The first derivative of y=(1+xf(x))/(sqrt(x)) is (2x^2\frac{df)/(dx)+xf(x)-1}{2xsqrt(x)}