What is the derivative of y=x4^{x+9} ?

The derivative of y=x4^{x+9} is 4^{x+9}+ln(2)*2^{2x+19}x

What is the first derivative of y=x4^{x+9} ?

The first derivative of y=x4^{x+9} is 4^{x+9}+ln(2)*2^{2x+19}x

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derivative of

y = x 4^{x + 9}

4^{x + 9} + ln (2) \cdot 2^{2 x + 19} x

Solution steps

\frac{d}{d x} (x \cdot 4^{x + 9})

Apply the Product Rule:

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

f = x, g = 4^{x + 9}

= \frac{d x}{d x} 4^{x + 9} + \frac{d}{d x} (4^{x + 9}) x

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

\frac{d}{d x} (4^{x + 9}) = ln (2) \cdot 2^{2 x + 19}

= 1 \cdot 4^{x + 9} + ln (2) \cdot 2^{2 x + 19} x

Simplify

= 4^{x + 9} + ln (2) \cdot 2^{2 x + 19} x

x \to \infty lim (- 2 e^{- x})

\int \frac{4}{3 x ^{2} 5 - 7 x ^{2}} d x

\frac{d}{d x} (\frac{4 x}{3 - cot ( x )})

f (x) = 6 ln (x)

What is the derivative of y=x4^{x+9} ?
The derivative of y=x4^{x+9} is 4^{x+9}+ln(2)*2^{2x+19}x
What is the first derivative of y=x4^{x+9} ?
The first derivative of y=x4^{x+9} is 4^{x+9}+ln(2)*2^{2x+19}x