what is the derivative of (ln(3))^cosx
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Originally Posted by jojoferni244 what is the derivative of (ln(3))^cosx recall that for any constant $\displaystyle a>0$, we have $\displaystyle \frac d{dx}a^x = a^x \ln a$ now, apply that rule with the chain rule to find this derivative. the hard way is to use logarithmic differentiation
Originally Posted by jojoferni244 what is the derivative of (ln(3))^cosx Let $\displaystyle u = \cos{x}$ so $\displaystyle y = \ln{3}^u$ $\displaystyle \frac{du}{dx} = -\sin{x}$ and $\displaystyle \frac{dy}{du} = \ln{3}^u\ln{\ln{3}} = \ln{3}^{\cos{x}}\ln{\ln{3}}$ $\displaystyle \frac{dy}{dx} = -\sin{x}\ln{3}^{\cos{x}}\ln{\ln{3}}$.
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