What is the derivative of (1/2)^x-x ?

The derivative of (1/2)^x-x is -ln(2)(1/2)^x-1

What is the first derivative of (1/2)^x-x ?

The first derivative of (1/2)^x-x is -ln(2)(1/2)^x-1

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derivative of

(\frac{1}{2})^{x} - x

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

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

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

\int_{- 1}^{1} \frac{3}{4} x^{2} (1 - x^{2}) d x

\int - 5 tan^{3} (x) d x

f (3)

y^{^{''}} + y = 9 e^{- 4 x}, y (0) = 0, y^{^{'}} (0) = 0

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

What is the derivative of (1/2)^x-x ?
The derivative of (1/2)^x-x is -ln(2)(1/2)^x-1
What is the first derivative of (1/2)^x-x ?
The first derivative of (1/2)^x-x is -ln(2)(1/2)^x-1