Find the derivative of the following functions.
f:x\longmapsto \cos\left(x\right)\cdot\log_2\left(x\right)
For a product of differentiable functions:
\dfrac{d}{dx} \left\{f\left(x\right)\cdot g\left(x\right)\right\} = f\left(x\right)\cdot g'\left(x\right) + f'\left(x\right)\cdot g\left(x\right).
We have:
\dfrac{d}{dx} \left\{ \cos{\left(x\right)} \right\} = -\sin{\left(x\right)}
And:
\dfrac{d}{dx}\left\{\log_2{\left(x\right)}\right\} =\dfrac{1}{x\ln{\left(2\right)}}
Thus, we have:
\dfrac{d}{dx}\left\{\cos{\left(x\right)}\log_2{\left(x\right)}\right\} = \cos{\left(x\right)}\cdot \dfrac{1}{x\ln{\left(2\right)}} + -\sin{\left(x\right)}\log_2{\left(x\right)}
For any x \in \mathbb{R^+_*} , f is differentiable and f'\left(x\right)=\dfrac{ \cos{\left(x\right)}}{x\ln{\left(2\right)}} -\sin{\left(x\right)}\log_2{\left(x\right)}.
f:x\longmapsto \sin{\left(x\right)}\cdot\cos{\left(x\right)}
For a product of differentiable functions:
\dfrac{d}{dx} \left\{f\left(x\right)\cdot g\left(x\right)\right\} = f\left(x\right)\cdot g'\left(x\right) + f'\left(x\right)\cdot g\left(x\right)
We have:
\dfrac{d}{dx} \left\{ \sin{\left(x\right)} \right\} = \cos{\left(x\right)}
And:
\dfrac{d}{dx}\left\{\cos{\left(x\right)}\right\} =-\sin{\left(x\right)}
Thus, we have:
\dfrac{d}{dx}\left\{\sin{\left(x\right)}\cos{\left(x\right)}\right\} = \sin{\left(x\right)}\left(-\sin{\left(x\right)}\right) + \cos{\left(x\right)}\cdot\cos{\left(x\right)}
For any x \in \mathbb{R} , f is differentiable and f'\left(x\right)=-\sin^2{\left(x\right)} + \cos^2{\left(x\right)}.
f:x\longmapsto 2^x\cdot\log_3{\left(x\right)}
For a product of differentiable functions:
\dfrac{d}{dx} \left\{f\left(x\right)\cdot g\left(x\right)\right\} = f\left(x\right)\cdot g'\left(x\right) + f'\left(x\right)\cdot g\left(x\right)
We have:
\dfrac{d}{dx} \left\{ 2^x \right\} = 2^x \ln{\left(2\right)}
And:
\dfrac{d}{dx}\left\{ \log_3{\left(x\right)} \right\} = \dfrac{1}{x\ln{\left(3\right)}}
Thus, we have:
\dfrac{d}{dx}\left\{2^x \log_3{\left(x\right)} \right\} = 2^x\left(\dfrac{1}{x\ln{\left(3\right)}}\right) + \log_3{\left(x\right)} \cdot 2^x\cdot \ln{\left(2\right)}
For any x \in \mathbb{R^+_*} , f is differentiable and f'\left(x\right)=2^x\left\{\dfrac{1}{x\ln{\left(3\right)}} + \ln{\left(2\right)}\log_3{\left(x\right)} \right\}.
f:x\longmapsto \ln{x}\cdot\left(x^4 + x^3 - \dfrac{1}{x}\right)
For a product of differentiable functions:
\dfrac{d}{dx} \left\{f\left(x\right)\cdot g\left(x\right)\right\} = f\left(x\right)\cdot g'\left(x\right) + f'\left(x\right)\cdot g\left(x\right)
We have:
\dfrac{d}{dx} \left\{ \ln{x} \right\} = \dfrac{1}{x}
And:
\dfrac{d}{dx}\left\{ x^4 + x^3 - \dfrac{1}{x} \right\} =4x^3 + 2x^2 + \dfrac{1}{x^2}
Thus, we have:
\dfrac{d}{dx}\left\{\ln{x}\left(x^4 + x^3 - \dfrac{1}{x}\right) \right\} = \ln{x}\left(4x^3 + 3x^2+ \dfrac{1}{x^2}\right) + \dfrac{1}{x}\left(x^4 + x^3 - \dfrac{1}{x}\right)
For any x \in \mathbb{R^+_*} , f is differentiable and f'\left(x\right)=\ln{x}\left(4x^3 + 3x^2 + \dfrac{1}{x^2}\right) + x^3 + x^2 - \dfrac{1}{x^2}.
f:x\longmapsto \log_4{\left(x\right)}\cdot\ln{\left(x\right)}
For a product of differentiable functions:
\dfrac{d}{dx} \left\{f\left(x\right)\cdot g\left(x\right)\right\} = f\left(x\right)\cdot g'\left(x\right) + f'\left(x\right)\cdot g\left(x\right)
We have:
\dfrac{d}{dx} \left\{ \log_4{\left(x\right)} \right\} = \dfrac{1}{x\ln{4}}
And:
\dfrac{d}{dx}\left\{\ln{x} \right\} =\dfrac{1}{x}
Thus, we have:
\dfrac{d}{dx}\left\{ \log_4{\left(x\right)} \ln{\left(x\right)} \right\} = \log_4{\left(x\right)}\cdot \dfrac{1}{x} + \dfrac{1}{x\ln{4}}\cdot \ln{\left(x\right)}
For any x \in \mathbb{R^+} , f is differentiable and f'\left(x\right) =\dfrac{1}{x}\left( \log_4{\left(x\right)} + \dfrac{\ln{\left(x\right)}}{\ln{4}}\right)
f:x\longmapsto 2^x.e^x
For a product of differentiable functions:
\dfrac{d}{dx} \left\{f\left(x\right)\cdot g\left(x\right)\right\} = f\left(x\right)\cdot g'\left(x\right) + f'\left(x\right)\cdot g\left(x\right)
We have:
\dfrac{d}{dx} \left\{ 2^x \right\} = 2^x \ln{2}
And:
\dfrac{d}{dx}\left\{e^x\right\} =e^x
Thus, we have:
\dfrac{d}{dx}\left\{2^x\cdot e^x\right\} = 2^x\cdot e^x + 2^x \cdot \ln{2} \cdot e^x
For any x \in \mathbb{R} , f is differentiable and f'\left(x\right) = 2^x\cdot e^x + 2^x \cdot e^x \cdot \ln{2}
f:x\longmapsto 2^x.\cos{\left(x\right)}
For a product of differentiable functions:
\dfrac{d}{dx} \left\{f\left(x\right)\cdot g\left(x\right)\right\} = f\left(x\right)\cdot g'\left(x\right) + f'\left(x\right)\cdot g\left(x\right)
We have:
\dfrac{d}{dx} \left\{ 2^x \right\} = 2^x\ln{2}
And:
\dfrac{d}{dx}\left\{\cos{x} \right\} =-\sin{x}
Thus, we have:
\dfrac{d}{dx}\left\{ 2^x\cos{\left(x\right)} \right\} = 2^x\left(-\sin{\left(x\right)}\right)+ \cos{\left(x\right)}\cdot2^x\cdot \ln{\left(2\right)}
For any x \in \mathbb{R} , f is differentiable and f'\left(x\right)= 2^x\left(\ln{2}\cdot \cos{x} -\sin{\left(x\right)}\right)