What is the derivative of (4x+3/(5e^x)) ?

The derivative of (4x+3/(5e^x)) is (-4x+1)/(5e^x)

What is the first derivative of (4x+3/(5e^x)) ?

The first derivative of (4x+3/(5e^x)) is (-4x+1)/(5e^x)

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\frac{d}{d x} (\frac{4 x + 3}{5 e ^{x}})

\frac{- 4 x + 1}{5 e ^{x}}

Solution steps

\frac{d}{d x} (\frac{4 x + 3}{5 e ^{x}})

Take the constant out:

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

= \frac{1}{5} \frac{d}{d x} (\frac{4 x + 3}{e ^{x}})

Apply the Quotient Rule:

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

= \frac{1}{5} \cdot \frac{\frac{d}{d x} ( 4 x + 3 ) e ^{x} - \frac{d}{d x} ( e ^{x} ) ( 4 x + 3 )}{( e ^{x} ) ^{2}}

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

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

= \frac{1}{5} \cdot \frac{4 e ^{x} - e ^{x} ( 4 x + 3 )}{( e ^{x} ) ^{2}}

Simplify

\frac{1}{5} \cdot \frac{4 e ^{x} - e ^{x} ( 4 x + 3 )}{( e ^{x} ) ^{2}} : \frac{- 4 x + 1}{5 e ^{x}}

= \frac{- 4 x + 1}{5 e ^{x}}

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

x \to - 19 lim (13 + x)

\int \frac{x ^{2}}{25 - x ^{2}} d x

\frac{d}{d x} (x^{y} + y^{x})

f (x) = x^{3} - 3 x + 2, at x = 3

What is the derivative of (4x+3/(5e^x)) ?
The derivative of (4x+3/(5e^x)) is (-4x+1)/(5e^x)
What is the first derivative of (4x+3/(5e^x)) ?
The first derivative of (4x+3/(5e^x)) is (-4x+1)/(5e^x)