Find the derivative of a function of the form x -> f(x)/g(x) - High school Calculus Exercises

Find the derivative of the following function.

f:x\longmapsto \dfrac{\cos\left(x\right)}{2^x}

f'\left(x\right)=\dfrac{-\sin{\left(x\right)}}{\ln{4^x}}

f'\left(x\right)=\dfrac{-\sin{\left(x\right)}}{2^x\ln{\left(2\right)}}

f'\left(x\right)= -\dfrac{\sin{\left(x\right)} + \cos{\left(x\right)} \cdot \ln{\left(2\right)}}{2^x}

f'\left(x\right)= -\dfrac{2^x\left(\sin{\left(x\right)} + \cos{\left(x\right)} \cdot \ln{\left(2\right)}\right)}{4^x}

\cos{\left(x\right)} is differentiable for all x \in \mathbb{R} and we have:

\dfrac{d}{dx}\left\{\cos{\left(x\right)}\right\} = -\sin{\left(x\right)}

2^x is differentiable for all x \in \mathbb{R} and we have:

\dfrac{d}{dx}\left\{2^x\right\} = 2^x \ln{2}.

For quotients of differentiable functions we have:

\dfrac{d}{dx}\left\{\dfrac{f\left(x\right)}{g\left(x\right)}\right\} = \dfrac{g\left(x\right)\cdot f'\left(x\right) - f\left(x\right)\cdot g'\left(x\right)}{g^2\left(x\right)}

Therefore:

\dfrac{d}{dx}\left\{\dfrac{\cos{\left(x\right)}}{2^x}\right\} = \dfrac{2^x\cdot \left(-\sin{\left(x\right)}\right) - \cos{\left(x\right)} \cdot 2^x\ln{\left(2\right)}}{\left[2^x\right]^2}

\dfrac{d}{dx}\left\{\dfrac{\cos{\left(x\right)}}{2^x}\right\} = \dfrac{-2^x\cdot \sin{\left(x\right)} - 2^x\cdot \cos{\left(x\right)} \cdot \ln{\left(2\right)}}{\left[2^x\right]^2}

For any x \in \mathbb{R} , f is differentiable and f'\left(x\right)= -\dfrac{\sin{\left(x\right)} + \cos{\left(x\right)} \cdot \ln{\left(2\right)}}{2^x}

f:x \longmapsto \dfrac{ \sin{\left(x\right)} } { \cos{\left(x\right)} }

f'\left(x\right)=\dfrac{\cos^2{\left(x\right)}-\sin^2{\left(x\right)}}{\cos^2{\left(x\right)}}

f'\left(x\right)=-\dfrac{\cos{\left(x\right)}} {\sin{\left(x\right)}}

f'\left(x\right)= \dfrac{1}{\cos^2{\left(x\right)}}

f'\left(x\right)= 1 + \sin^2{\left(x\right)}

\cos{\left(x\right)} is differentiable for all x \in \mathbb{R} and we have:

\dfrac{d}{dx}\left\{\cos{\left(x\right)}\right\} = -\sin{\left(x\right)}

\sin{\left(x\right)} is differentiable for all x \in \mathbb{R} and we have:

\dfrac{d}{dx}\left\{ \sin{\left(x\right)} \right\} = \cos{\left(x\right)}.

For quotients of differentiable functions we have:

\dfrac{d}{dx}\left\{\dfrac{f\left(x\right)}{g\left(x\right)}\right\} = \dfrac{g\left(x\right)\cdot f'\left(x\right) - f\left(x\right)\cdot g'\left(x\right)}{g^2\left(x\right)}

Therefore:

\dfrac{d}{dx}\left\{\dfrac{\sin{\left(x\right)}}{\cos{\left(x\right)}}\right\} = \dfrac{\cos{\left(x\right)}\cdot\cos{\left(x\right)} - \sin{\left(x\right)}\cdot\left(-\sin{\left(x\right)}\right) }{\cos^2{\left(x\right)}}

\dfrac{d}{dx}\left\{\dfrac{\sin{\left(x\right)}}{\cos{\left(x\right)}}\right\} =\dfrac{\cos^2{\left(x\right)} + \sin^2{\left(x\right)} } {\cos^2{\left(x\right)}}

\dfrac{d}{dx}\left\{\dfrac{\sin{\left(x\right)}}{\cos{\left(x\right)}}\right\} =\dfrac{1} {\cos^2{\left(x\right)}}

For any x \in \mathbb{R} , f is differentiable and f'\left(x\right)= \dfrac{1}{\cos^2{\left(x\right)}}

f:x \longmapsto \dfrac{ \log_3{\left(x\right)} } {\sin{\left(x\right)}}

f'\left(x\right) = \dfrac{\sin\left(x\right)- x\ln\left(x\right)\cos{\left(x\right)}}{x\cdot\sin^2{\left(x\right)} \cdot \ln{3}}

f'\left(x\right) = \dfrac{x\ln{3}}{\sin{\left(x\right)}} + \dfrac{\log_3{\left(x\right)}\cdot \cos{\left(x\right)}}{\sin^2{\left(x\right)}}

f'\left(x\right) = \dfrac{1}{x\cdot\sin{\left(x\right)} \cdot \ln{3}} - \log_3{\left(x\right)}\cdot \cos{\left(x\right)}

f'\left(x\right)= \dfrac{x\ln{3}}{\cos{\left(x\right)}}

\log_3{\left(x\right)} is differentiable for all x \in \mathbb{R}^+_* and we have:

\dfrac{d}{dx}\left\{\log_3{\left(x\right)} \right\} = \dfrac{1}{x\ln{3}}

\sin{\left(x\right)} is differentiable for all x \in \mathbb{R} and we have:

\dfrac{d}{dx}\left\{ \sin{\left(x\right)} \right\} = \cos{\left(x\right)}.

For quotients of differentiable functions we have:

\dfrac{d}{dx}\left\{\dfrac{f\left(x\right)}{g\left(x\right)}\right\} = \dfrac{g\left(x\right)\cdot f'\left(x\right) - f\left(x\right)\cdot g'\left(x\right)}{g^2\left(x\right)}

Therefore:

\dfrac{d}{dx}\left\{\dfrac{\log_3{\left(x\right)}}{\sin{\left(x\right)}}\right\} = \dfrac{ \sin{\left(x\right)}\left(\dfrac{1}{x\ln{3}}\right) - \log_3{\left(x\right)} \cdot \cos{\left(x\right)}}{\sin^2{\left(x\right)}}

\dfrac{d}{dx}\left\{\dfrac{\log_3{\left(x\right)}}{\sin\left(x\right)}\right\} = \dfrac{\sin\left(x\right)- \log_3{\left(x\right)\ln\left(3\right)}\cos{\left(x\right)}}{x\cdot\sin^2{\left(x\right)} \cdot \ln{3}}

The logarithm is defined as:

\log_3\left(x\right) = \dfrac{\ln\left(x\right)}{\ln\left(3\right)}

So:

\dfrac{d}{dx}\left\{\dfrac{\log_3{\left(x\right)}}{\sin\left(x\right)}\right\} = \dfrac{\sin\left(x\right)- x\ln\left(x\right)\cos{\left(x\right)}}{x\cdot\sin^2{\left(x\right)} \cdot \ln{3}}

For any x \in \mathbb{R}^+_* , f is differentiable and f'\left(x\right) = \dfrac{\sin\left(x\right)- x\ln\left(x\right)\cos{\left(x\right)}}{x\cdot\sin^2{\left(x\right)} \cdot \ln{3}}

f:x \longmapsto \dfrac{e^{x} } {3^x}

f'\left(x\right) = \dfrac{ e^x }{3^{x}}

f'\left(x\right) = \dfrac{ e^x \left(1 - \ln{3}\right)}{3^{x}}

f'\left(x\right) = \dfrac{ e^x \left(1 - \ln{3}\right)}{2}

f'\left(x\right) = -\dfrac{ e^x \left(1 + \ln{3}\right)}{3^{x}}

e^x is differentiable for all x \in \mathbb{R} and we have:

\dfrac{d}{dx}\left\{ e^x \right\} = e^x

3^x is differentiable for all x \in \mathbb{R} and we have:

\dfrac{d}{dx}\left\{ 3^x \right\} = 3^x\ln{3}.

For quotients of differentiable functions we have:

\dfrac{d}{dx}\left\{\dfrac{f\left(x\right)}{g\left(x\right)}\right\} = \dfrac{g\left(x\right)\cdot f'\left(x\right) - f\left(x\right)\cdot g'\left(x\right)}{g^2\left(x\right)}

Therefore:

\dfrac{d}{dx}\left\{\dfrac{e^x}{3^x}\right\} = \dfrac{3^x\cdot e^x - e^x \cdot 3^x \ln{3}}{\left(3^x\right)^2}

\dfrac{d}{dx}\left\{\dfrac{e^x}{3^x}\right\} = \dfrac{3^x\cdot e^x \left(1 - \ln{3}\right)}{3^{2x}}

\dfrac{d}{dx}\left\{\dfrac{e^x}{3^x}\right\} = \dfrac{ e^x \left(1 - \ln{3}\right)}{3^{x}}

For any x \in \mathbb{R} , f is differentiable and f'\left(x\right) = \dfrac{ e^x \left(1 - \ln{3}\right)}{3^{x}}

f:x \longmapsto \dfrac{\ln{x} } {\log_2{x}}

f'\left(x\right) = \ln{2}

f'\left(x\right) = 1-\dfrac{1}{ \ln{2}\cdot \ln{\left(x\right)}\cdot \log_2{\left(x\right)} }

f'\left(x\right) = -\dfrac{1}{\ln{2}}

f'\left(x\right) = 0

\log_2{\left(x\right)} can be expressed as a function of \ln\left(x\right) :

\log_2\left(x\right) = \dfrac{\ln x}{\ln 2}

Thus:

\dfrac{\ln\left(x\right)}{\log_2\left(x\right)} = \ln 2

Therefore:

\dfrac{d}{dx}\left\{\dfrac{\ln{x}}{\log_2{\left(x\right)}}\right\} = 0

For any x \in \mathbb{R^+_*} , f is differentiable and f'\left(x\right) = 0

f:x \longmapsto \dfrac{\ln{x} } {4^x}

f'\left(x\right) =\dfrac{1}{4^x} \left(\dfrac{1}{x} - \ln{4x}\right)

f'\left(x\right) =\dfrac{1}{4^x} \left(\dfrac{1}{x} - \ln{x} \cdot \ln{4}\right)

f'\left(x\right) =\dfrac{1}{4} - \ln{x} \cdot \ln{4}

f'\left(x\right) = \dfrac{1}{ x}\cdot\dfrac{1}{4^x\ln{4}}

\ln{\left(x\right)} is differentiable for all x \in \mathbb{R^+} and we have:

\dfrac{d}{dx}\left\{ \ln{\left(x\right)} \right\} = \dfrac{1}{x}

4^x is differentiable for all x \in \mathbb{R} and we have:

\dfrac{d}{dx}\left\{ 4^x \right\} = 4^x\ln{4}.

For quotients of differentiable functions we have:

\dfrac{d}{dx}\left\{\dfrac{f\left(x\right)}{g\left(x\right)}\right\} = \dfrac{g\left(x\right)\cdot f'\left(x\right) - f\left(x\right)\cdot g'\left(x\right)}{g^2\left(x\right)}

Therefore:

\dfrac{d}{dx}\left\{\dfrac{\ln{x}}{4^x}\right\} = \dfrac{4^x\left(\dfrac{1}{x}\right) - \ln{\left(x\right)} \cdot 4^x \cdot \ln{4}}{\left(4^x\right)^2}

\dfrac{d}{dx}\left\{\dfrac{\ln{x}}{4^x}\right\} = \dfrac{\left(\dfrac{1}{x}\right) - \ln{\left(x\right)} \cdot \ln{4}}{4^x}

\dfrac{d}{dx}\left\{\dfrac{\ln{x}}{4^x}\right\} =\dfrac{1}{4^x} \left(\dfrac{1}{x} - \ln{x} \cdot \ln{4}\right)

For any x \in \mathbb{R^+} , f is differentiable and f'\left(x\right) =\dfrac{1}{4^x} \left(\dfrac{1}{x} - \ln{x} \cdot \ln{4}\right)

f:x \longmapsto \dfrac{1/x } {\ln{\left(x\right)}}

f'\left(x\right)= -\dfrac{1}{\ln{\left(x^2\right)}}

f'\left(x\right)= \dfrac{1}{x^2}\left(\dfrac{1}{ \ln{\left(x\right)}}\right)

f'\left(x\right)= -\dfrac{1}{x^2}\left(\dfrac{1}{\ln{\left(x\right)}}\right)

f'\left(x\right)= -\dfrac{1}{x^2}\left(\dfrac{ \ln{\left(x\right)} +1}{\ln^2{\left(x\right)}}\right)

\dfrac{1}{x} is differentiable for all x \neq 0 and we have:

\dfrac{d}{dx}\left\{ \dfrac{1}{x} \right\} = -\dfrac{1}{x^2}.

\ln{\left(x\right)} is differentiable for all x \in \mathbb{R^+} and we have:

\dfrac{d}{dx}\left\{ \ln{\left(x\right)} \right\} = \dfrac{1}{x}

For quotients of differentiable functions we have:

\dfrac{d}{dx}\left\{\dfrac{f\left(x\right)}{g\left(x\right)}\right\} = \dfrac{g\left(x\right)\cdot f'\left(x\right) - f\left(x\right)\cdot g'\left(x\right)}{g^2\left(x\right)}

Therefore:

\dfrac{d}{dx}\left\{\dfrac{1/x}{\ln{\left(x\right)}}\right\} = \dfrac{ \ln{\left(x\right)}\left(-\dfrac{1}{x^2}\right) - \dfrac{1}{x}\left(\dfrac{1}{x}\right)}{\ln^2{\left(x\right)}}

\dfrac{d}{dx}\left\{\dfrac{1/x}{\ln{\left(x\right)}}\right\} = -\dfrac{1}{x^2}\left(\dfrac{ \ln{\left(x\right)} +1}{\ln^2{\left(x\right)}}\right)

For any x \in \mathbb{R^+} , f is differentiable and f'\left(x\right)= -\dfrac{1}{x^2}\left(\dfrac{ \ln{\left(x\right)} +1}{\ln^2{\left(x\right)}}\right)