What is (\partial)/(\partial x)(10^{x^2-1}) ?

The answer to (\partial)/(\partial x)(10^{x^2-1}) is ln(10)*2^{x^2}*5^{x^2-1}x

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(\partial)/(\partial x)(10^{x^2-1})

\frac{\partial}{\partial x} (1 0^{x^{2} - 1})

ln (10) \cdot 2^{x^{2}} \cdot 5^{x^{2} - 1} x

Solution steps

\frac{\partial}{\partial x} (1 0^{x^{2} - 1})

Apply exponent rule:

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

1 0^{x^{2} - 1} = e^{(x^{2} - 1) l n (10)}

= \frac{\partial}{\partial x} (e^{(x^{2} - 1) l n (10)})

Apply the chain rule:

e^{(x^{2} - 1) l n (10)} \frac{\partial}{\partial x} ((x^{2} - 1) ln (10))

= e^{(x^{2} - 1) l n (10)} \frac{\partial}{\partial x} ((x^{2} - 1) ln (10))

\frac{\partial}{\partial x} ((x^{2} - 1) ln (10)) = 2 ln (10) x

= e^{(x^{2} - 1) l n (10)} \cdot 2 ln (10) x

Simplify

e^{(x^{2} - 1) l n (10)} \cdot 2 ln (10) x : ln (10) \cdot 2^{x^{2}} \cdot 5^{x^{2} - 1} x

= ln (10) \cdot 2^{x^{2}} \cdot 5^{x^{2} - 1} x

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What is (\partial)/(\partial x)(10^{x^2-1}) ?
The answer to (\partial)/(\partial x)(10^{x^2-1}) is ln(10)*2^{x^2}*5^{x^2-1}x