What is the derivative of (7x^2/(4e^x-x)) ?

The derivative of (7x^2/(4e^x-x)) is (7(-x^2-4e^xx^2+8e^xx))/((4e^x-x)^2)

What is the first derivative of (7x^2/(4e^x-x)) ?

The first derivative of (7x^2/(4e^x-x)) is (7(-x^2-4e^xx^2+8e^xx))/((4e^x-x)^2)

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\frac{d}{d x} (\frac{7 x ^{2}}{4 e ^{x} - x})

\frac{7 ( - x ^{2} - 4 e ^{x} x ^{2} + 8 e ^{x} x )}{( 4 e ^{x} - x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{7 x ^{2}}{4 e ^{x} - x})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 7 \frac{d}{d x} (\frac{x ^{2}}{4 e ^{x} - x})

Apply the Quotient Rule:

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

= 7 \cdot \frac{\frac{d}{d x} ( x ^{2} ) ( 4 e ^{x} - x ) - \frac{d}{d x} ( 4 e ^{x} - x ) x ^{2}}{( 4 e ^{x} - x ) ^{2}}

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

\frac{d}{d x} (4 e^{x} - x) = 4 e^{x} - 1

= 7 \cdot \frac{2 x ( 4 e ^{x} - x ) - ( 4 e ^{x} - 1 ) x ^{2}}{( 4 e ^{x} - x ) ^{2}}

Simplify

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

= \frac{7 ( - x ^{2} - 4 e ^{x} x ^{2} + 8 e ^{x} x )}{( 4 e ^{x} - x ) ^{2}}

\frac{\partial}{\partial y} (e^{x y z^{3}} y z^{3})

y = - \frac{1}{4} x + 5

f (x) = 3 cos (x), at x = \frac{π}{4}

\int 4 cos^{3} (4 x) \cdot sin (4 x) d x

\frac{( x ^{2} + 1 ) d y}{d x} + 9 x (y - 1) = 0

What is the derivative of (7x^2/(4e^x-x)) ?
The derivative of (7x^2/(4e^x-x)) is (7(-x^2-4e^xx^2+8e^xx))/((4e^x-x)^2)
What is the first derivative of (7x^2/(4e^x-x)) ?
The first derivative of (7x^2/(4e^x-x)) is (7(-x^2-4e^xx^2+8e^xx))/((4e^x-x)^2)