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endeavor
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Could someone please make sure I'm doing this right.
I want to find the derivative of the logarithm to the base a of x, using implicit differentiation.
Let y = \log_{a} x
a^y = x
\frac{d}{dx} (a^y) = 1 (implicit differentiation)
\frac{d}{dx} (e^{\ln a})^y = 1
\frac{d}{dx} (e^{(\ln a)y}) = 1
e^{(\ln a)y} \frac{d}{dx} ((\ln a)(y)) = 1
(e^{\ln a})^y \frac{d}{dy} ((\ln a)(y)) \frac{dy}{dx} = 1 (did I use the chain rule correctly here?)
(a^y)(\ln a) \frac{dy}{dx} = 1
x \ln a \frac{dy}{dx} = 1
\frac{dy}{dx} = \frac{1}{x \ln a}
I want to find the derivative of the logarithm to the base a of x, using implicit differentiation.
Let y = \log_{a} x
a^y = x
\frac{d}{dx} (a^y) = 1 (implicit differentiation)
\frac{d}{dx} (e^{\ln a})^y = 1
\frac{d}{dx} (e^{(\ln a)y}) = 1
e^{(\ln a)y} \frac{d}{dx} ((\ln a)(y)) = 1
(e^{\ln a})^y \frac{d}{dy} ((\ln a)(y)) \frac{dy}{dx} = 1 (did I use the chain rule correctly here?)
(a^y)(\ln a) \frac{dy}{dx} = 1
x \ln a \frac{dy}{dx} = 1
\frac{dy}{dx} = \frac{1}{x \ln a}
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