What is the derivative of 2^{sqrt(x^2+3x+4)} ?

The derivative of 2^{sqrt(x^2+3x+4)} is (ln(2)*2^{sqrt(x^2+3x+4)-1}(2x+3))/(sqrt(x^2+3x+4))

What is the first derivative of 2^{sqrt(x^2+3x+4)} ?

The first derivative of 2^{sqrt(x^2+3x+4)} is (ln(2)*2^{sqrt(x^2+3x+4)-1}(2x+3))/(sqrt(x^2+3x+4))

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\frac{d}{d x} (2^{x^{2} + 3 x + 4})

\frac{ln ( 2 ) \cdot 2 ^{x^{2} + 3 x + 4 - 1} ( 2 x + 3 )}{x ^{2} + 3 x + 4}

Solution steps

\frac{d}{d x} (2^{x^{2} + 3 x + 4})

Apply exponent rule:

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

2^{x^{2} + 3 x + 4} = e^{x^{2} + 3 x + 4 l n (2)}

= \frac{d}{d x} (e^{x^{2} + 3 x + 4 l n (2)})

Apply the chain rule:

e^{x^{2} + 3 x + 4 l n (2)} \frac{d}{d x} (x^{2} + 3 x + 4 ln (2))

= e^{x^{2} + 3 x + 4 l n (2)} \frac{d}{d x} (x^{2} + 3 x + 4 ln (2))

\frac{d}{d x} (x^{2} + 3 x + 4 ln (2)) = \frac{ln ( 2 ) ( 2 x + 3 )}{2 x ^{2} + 3 x + 4}

= e^{x^{2} + 3 x + 4 l n (2)} \frac{ln ( 2 ) ( 2 x + 3 )}{2 x ^{2} + 3 x + 4}

Simplify

e^{x^{2} + 3 x + 4 l n (2)} \frac{ln ( 2 ) ( 2 x + 3 )}{2 x ^{2} + 3 x + 4} : \frac{ln ( 2 ) \cdot 2 ^{x^{2} + 3 x + 4 - 1} ( 2 x + 3 )}{x ^{2} + 3 x + 4}

= \frac{ln ( 2 ) \cdot 2 ^{x^{2} + 3 x + 4 - 1} ( 2 x + 3 )}{x ^{2} + 3 x + 4}

x^{3} - 3 x

\frac{d}{d x} (x 9^{x})

\frac{d}{d x} (a x^{a - 1} ln (x) + x^{a - 1})

- \frac{6 e ^{\frac{6}{x}}}{x ^{2}}

\frac{d}{d x} (\frac{5 x ^{2}}{2 b ^{2} + x ^{2}})

What is the derivative of 2^{sqrt(x^2+3x+4)} ?
The derivative of 2^{sqrt(x^2+3x+4)} is (ln(2)*2^{sqrt(x^2+3x+4)-1}(2x+3))/(sqrt(x^2+3x+4))
What is the first derivative of 2^{sqrt(x^2+3x+4)} ?
The first derivative of 2^{sqrt(x^2+3x+4)} is (ln(2)*2^{sqrt(x^2+3x+4)-1}(2x+3))/(sqrt(x^2+3x+4))