What is the derivative of f(x)=5^{-x/2}sin(2x) ?

The derivative of f(x)=5^{-x/2}sin(2x) is -(ln(5)e^{-(ln(5)x)/2}sin(2x))/(2)+2*5^{-x/2}cos(2x)

What is the first derivative of f(x)=5^{-x/2}sin(2x) ?

The first derivative of f(x)=5^{-x/2}sin(2x) is -(ln(5)e^{-(ln(5)x)/2}sin(2x))/(2)+2*5^{-x/2}cos(2x)

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derivative of

f (x) = 5^{- \frac{x}{2}} sin (2 x)

- \frac{ln ( 5 ) e ^{- \frac{l n ( 5 ) x}{2}} sin ( 2 x )}{2} + 2 \cdot 5^{- \frac{x}{2}} cos (2 x)

Solution steps

\frac{d}{d x} (5^{- \frac{x}{2}} sin (2 x))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = 5^{- \frac{x}{2}}, g = sin (2 x)

= \frac{d}{d x} (5^{- \frac{x}{2}}) sin (2 x) + \frac{d}{d x} (sin (2 x)) 5^{- \frac{x}{2}}

\frac{d}{d x} (5^{- \frac{x}{2}}) = - \frac{ln ( 5 ) e ^{- \frac{l n ( 5 ) x}{2}}}{2}

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

= (- \frac{ln ( 5 ) e ^{- \frac{l n ( 5 ) x}{2}}}{2}) sin (2 x) + cos (2 x) \cdot 2 \cdot 5^{- \frac{x}{2}}

Simplify

(- \frac{ln ( 5 ) e ^{- \frac{l n ( 5 ) x}{2}}}{2}) sin (2 x) + cos (2 x) \cdot 2 \cdot 5^{- \frac{x}{2}} : - \frac{ln ( 5 ) e ^{- \frac{l n ( 5 ) x}{2}} sin ( 2 x )}{2} + 2 \cdot 5^{- \frac{x}{2}} cos (2 x)

= - \frac{ln ( 5 ) e ^{- \frac{l n ( 5 ) x}{2}} sin ( 2 x )}{2} + 2 \cdot 5^{- \frac{x}{2}} cos (2 x)

x \to 0 lim (x^{4} + 2 e^{x} - 2)

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

\int_{1}^{3} r^{3} ln (r) d r

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

7 x^{2} - x^{3}

What is the derivative of f(x)=5^{-x/2}sin(2x) ?
The derivative of f(x)=5^{-x/2}sin(2x) is -(ln(5)e^{-(ln(5)x)/2}sin(2x))/(2)+2*5^{-x/2}cos(2x)
What is the first derivative of f(x)=5^{-x/2}sin(2x) ?
The first derivative of f(x)=5^{-x/2}sin(2x) is -(ln(5)e^{-(ln(5)x)/2}sin(2x))/(2)+2*5^{-x/2}cos(2x)