What is the derivative of 2^xx^2 ?

The derivative of 2^xx^2 is ln(2)*2^xx^2+2^{1+x}x

What is the first derivative of 2^xx^2 ?

The first derivative of 2^xx^2 is ln(2)*2^xx^2+2^{1+x}x

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

ln (2) \cdot 2^{x} x^{2} + 2^{1 + x} x

Solution steps

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

Apply the Product Rule:

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

f = 2^{x}, g = x^{2}

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

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

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

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

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

2 \cdot 2^{x} = 2^{1 + x}

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

\int \frac{1}{8 - 7 x} d x

\int (x^{2} + 2 x) cos (x) d x

\frac{d}{d x} ((x^{4}))

y^{^{''}} + 4 y = 5 sin (2 x)

\int_{7}^{\infty} x e^{- 4 x} d x

What is the derivative of 2^xx^2 ?
The derivative of 2^xx^2 is ln(2)*2^xx^2+2^{1+x}x
What is the first derivative of 2^xx^2 ?
The first derivative of 2^xx^2 is ln(2)*2^xx^2+2^{1+x}x