What is the derivative of y=(3+7x-8sqrt(x))/x ?

The derivative of y=(3+7x-8sqrt(x))/x is (4sqrt(x)-3)/(x^2)

What is the first derivative of y=(3+7x-8sqrt(x))/x ?

The first derivative of y=(3+7x-8sqrt(x))/x is (4sqrt(x)-3)/(x^2)

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derivative of

y = \frac{3 + 7 x - 8 x}{x}

\frac{4 x - 3}{x ^{2}}

Solution steps

\frac{d}{d x} (\frac{3 + 7 x - 8 x}{x})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d x} ( 3 + 7 x - 8 x ) x - \frac{d x}{d x} ( 3 + 7 x - 8 x )}{x ^{2}}

\frac{d}{d x} (3 + 7 x - 8 x) = - \frac{4}{x} + 7

\frac{d x}{d x} = 1

= \frac{( - \frac{4}{x} + 7 ) x - 1 \cdot ( 3 + 7 x - 8 x )}{x ^{2}}

Simplify

\frac{( - \frac{4}{x} + 7 ) x - 1 \cdot ( 3 + 7 x - 8 x )}{x ^{2}} : \frac{4 x - 3}{x ^{2}}

= \frac{4 x - 3}{x ^{2}}

(x^{2} e^{x + 3})^{^{'}}

y^{^{'}} = - 3 y + e^{- 2 x}, y (0) = 0

f (x) = \frac{2}{1 + x ^{2}}, h (x) = x^{2}

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

\frac{d ^{2}}{d x ^{2}} (ln (x))

What is the derivative of y=(3+7x-8sqrt(x))/x ?
The derivative of y=(3+7x-8sqrt(x))/x is (4sqrt(x)-3)/(x^2)
What is the first derivative of y=(3+7x-8sqrt(x))/x ?
The first derivative of y=(3+7x-8sqrt(x))/x is (4sqrt(x)-3)/(x^2)