What is the derivative of e^x+2^{(-x}+2cos(x)-6) ?

The derivative of e^x+2^{(-x}+2cos(x)-6) is e^x-ln(2)*2^{-x}-2sin(x)

What is the first derivative of e^x+2^{(-x}+2cos(x)-6) ?

The first derivative of e^x+2^{(-x}+2cos(x)-6) is e^x-ln(2)*2^{-x}-2sin(x)

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\frac{d}{d x} (e^{x} + 2^{(- x)} + 2 cos (x) - 6)

e^{x} - ln (2) \cdot 2^{- x} - 2 sin (x)

Solution steps

\frac{d}{d x} (e^{x} + 2^{(- x)} + 2 cos (x) - 6)

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (e^{x}) + \frac{d}{d x} (2^{- x}) + \frac{d}{d x} (2 cos (x)) - \frac{d}{d x} (6)

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

\frac{d}{d x} (2^{- x}) = - ln (2) \cdot 2^{- x}

\frac{d}{d x} (2 cos (x)) = - 2 sin (x)

\frac{d}{d x} (6) = 0

= e^{x} - ln (2) \cdot 2^{- x} - 2 sin (x) - 0

Simplify

= e^{x} - ln (2) \cdot 2^{- x} - 2 sin (x)

\frac{d y}{d x} + 8 y = e^{x} y^{- 6}

\frac{2 x d y}{d x} + y = 6 x

\frac{d}{d x} (4 x^{3} - 16 x + 5)

\int \frac{1}{1 + x ^{2} + y ^{2}} d y

y^{^{''}} + 16 y = 2 sin (4 t)

What is the derivative of e^x+2^{(-x}+2cos(x)-6) ?
The derivative of e^x+2^{(-x}+2cos(x)-6) is e^x-ln(2)*2^{-x}-2sin(x)
What is the first derivative of e^x+2^{(-x}+2cos(x)-6) ?
The first derivative of e^x+2^{(-x}+2cos(x)-6) is e^x-ln(2)*2^{-x}-2sin(x)