derivative of (e^x-e^{-x}/5) — step-by-step solution

⚠ (e^x-e^{-x}/5) is read as (e^x-e^(-x)/5): use ^(...) to group an exponent, use * to multiply

Let's perform the differentiation:

\frac{d}{d x} (e^{x} - \frac{e^{- x}}{5})

ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (e^{x} - \frac{e^{- x}}{5})

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

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

ⓘ The derivative of exp(x) is exp(x):

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

e^{x}

■A constant factor stays in front of the derivative, (k*u)' = k*u'. The derivative of exp(x) is exp(x):

\frac{d}{d x} (- \frac{e^{- x}}{5})

- \frac{1}{5} \cdot \frac{d}{d x} (- x) \cdot e^{- x}

e^{x} - \frac{1}{5} \cdot \frac{d}{d x} (- x) \cdot e^{- x}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{1}{5} \cdot \frac{d}{d x} (- x) \cdot e^{- x}

\frac{1 \cdot \frac{d}{d x} (- x) \cdot e^{- x}}{5}

\frac{\frac{d}{d x} (- x) \cdot e^{- x}}{5}

■A negative exponent moves the power to the other side of the fraction bar, for a nonzero base:

- \frac{\frac{d}{d x} (- x) \cdot e^{- x}}{5}

- \frac{\frac{d}{d x} (- x)}{e^{x} \cdot 5}

e^{x} - \frac{\frac{d}{d x} (- x)}{e^{x} \cdot 5}

ⓘ The derivative of x is 1:

\frac{d}{d x} (- x)

-1

e^{x} - (\frac{-1}{e^{x} \cdot 5})

ⓘ Two minus signs cancel out:

- (\frac{-1}{e^{x} \cdot 5})

\frac{1}{e^{x} \cdot 5}

e^{x} + \frac{1}{e^{x} \cdot 5}

▶Derivative:

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