What is the derivative of 6^{x^2-3} ?

The derivative of 6^{x^2-3} is ln(6)*3^{x^2-3}*2^{x^2-2}x

What is the first derivative of 6^{x^2-3} ?

The first derivative of 6^{x^2-3} is ln(6)*3^{x^2-3}*2^{x^2-2}x

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\frac{d}{d x} (6^{x^{2} - 3})

ln (6) \cdot 3^{x^{2} - 3} \cdot 2^{x^{2} - 2} x

Solution steps

\frac{d}{d x} (6^{x^{2} - 3})

Apply exponent rule:

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

6^{x^{2} - 3} = e^{(x^{2} - 3) l n (6)}

= \frac{d}{d x} (e^{(x^{2} - 3) l n (6)})

Apply the chain rule:

e^{(x^{2} - 3) l n (6)} \frac{d}{d x} ((x^{2} - 3) ln (6))

= e^{(x^{2} - 3) l n (6)} \frac{d}{d x} ((x^{2} - 3) ln (6))

\frac{d}{d x} ((x^{2} - 3) ln (6)) = 2 ln (6) x

= e^{(x^{2} - 3) l n (6)} \cdot 2 ln (6) x

Simplify

e^{(x^{2} - 3) l n (6)} \cdot 2 ln (6) x : ln (6) \cdot 3^{x^{2} - 3} \cdot 2^{x^{2} - 2} x

= ln (6) \cdot 3^{x^{2} - 3} \cdot 2^{x^{2} - 2} x

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

\int \frac{x ^{2}}{x ^{2} - 6 x + 9} d x

d er i v a t i v e \frac{2 x ^{2} - 7}{8 x ^{3} + 1}

\frac{d}{d x} (\frac{x ^{2} + y ^{2} - 9}{x})

n = 1 \sum \infty \frac{- 1 ^{n + 1}}{n}

What is the derivative of 6^{x^2-3} ?
The derivative of 6^{x^2-3} is ln(6)*3^{x^2-3}*2^{x^2-2}x
What is the first derivative of 6^{x^2-3} ?
The first derivative of 6^{x^2-3} is ln(6)*3^{x^2-3}*2^{x^2-2}x