What is the derivative of g(x)=2^{x^2+1x} ?

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

What is the first derivative of g(x)=2^{x^2+1x} ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

g (x) = 2^{x^{2} + 1 x}

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

Solution steps

\frac{d}{d x} (2^{x^{2} + 1 \cdot x})

Simplify

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

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

Apply exponent rule:

a^{b} = e^{b l n (a)}

2^{x^{2} + x} = e^{(x^{2} + x) l n (2)}

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

Apply the chain rule:

e^{(x^{2} + x) l n (2)} \frac{d}{d x} ((x^{2} + x) ln (2))

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

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

= e^{(x^{2} + x) l n (2)} ln (2) (2 x + 1)

e^{(x^{2} + x) l n (2)} = 2^{x^{2} + x}

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

\frac{\partial}{\partial x} (x^{2} tan (4 x^{4} y))

\frac{d}{d x} (x^{2} sin^{3} (5 x))

\int \frac{1}{x ^{2} x ^{2} - 64} d x

\int \frac{1}{t ^{2} - 8 t + 25} d t

\int e^{2 x + 5} d x

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