Find the derivative of a function of the form x -> a.f(x)+g(x) Calculus

We have:

\dfrac{d}{dx}\left\{3\sin{\left(x\right)}\right\} = 3\dfrac{d}{dx}\left\{\sin{\left(x\right)}\right\} = 3\cos{\left(x\right)}

We also have:

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

For a linear combination of differentiable functions:

\dfrac{d}{dx}\left\{f\left(x\right) + g\left(x\right)\right\} = f'\left(x\right) + g'\left(x\right)

Therefore:

\dfrac{d}{dx}\left\{3\sin{\left(x\right)} + 4^x\right\} = 3\cos{\left(x\right)} + 4^x\left(\ln{4}\right)

We have:

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

We also have:

\dfrac{d}{dx}\left\{\log_4{\left(x\right)} \right\} = \dfrac{1}{x\ln{4}}

For a linear combination of differentiable functions:

\dfrac{d}{dx}\left\{f\left(x\right) + g\left(x\right)\right\} = f'\left(x\right) + g'\left(x\right)

Therefore:

\dfrac{d}{dx}\left\{-2\cos{\left(x\right)} + \log_4{\left(x\right)} \right\} = 2\sin{\left(x\right)} + \dfrac{1}{x\ln{4}}

We have:

\dfrac{d}{dx}\left\{-4\cdot 3^x\right\} = -4\dfrac{d}{dx}\left\{3^x\right\} = -4\left(3^x\ln{3}\right) = -4\left(3^x\right)\ln{3}

We also have:

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

For a linear combination of differentiable functions:

\dfrac{d}{dx}\left\{f\left(x\right) + g\left(x\right)\right\} = f'\left(x\right) + g'\left(x\right)

Therefore:

\dfrac{d}{dx}\left\{-4\cdot3^x + \cos{\left(x\right)}\right\} = -4\left(3^x\right)\ln{\left(3\right)} - \sin{\left(x\right)}

We have:

\dfrac{d}{dx}\left\{-5\log_2{\left(x\right)}\right\} = -5\dfrac{d}{dx}\left\{\log_2{\left(x\right)}\right\} = -5\left(\dfrac{1}{x\ln{2}}\right)

We also have:

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

For a linear combination of differentiable functions:

\dfrac{d}{dx}\left\{f\left(x\right) + g\left(x\right)\right\} = f'\left(x\right) + g'\left(x\right)

Therefore:

\dfrac{d}{dx}\left\{-5\log_2{\left(x\right)} + \cos{\left(x\right)} \right\} = \dfrac{-5}{x\ln{2}} -\sin{\left(x\right)}

We have:

\dfrac{d}{dx}\left\{-3e^x\right\} = 3\dfrac{d}{dx}\left\{e^x\right\} = 3e^x

We also have:

\dfrac{d}{dx}\left\{5^x \right\} = 5^x\ln{5}

For a linear combination of differentiable functions:

\dfrac{d}{dx}\left\{f\left(x\right) + g\left(x\right)\right\} = f'\left(x\right) + g'\left(x\right)

Therefore:

\dfrac{d}{dx}\left\{ 3e^x + 5^x\right\} = 3e^x + 5^x\ln{5}

We have:

\dfrac{d}{dx}\left\{2\cdot6^x\right\} = 2\dfrac{d}{dx}\left\{6^x\right\} = 2\cdot 6^x \cdot \ln{6}

We also have:

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

For a linear combination of differentiable functions:

\dfrac{d}{dx}\left\{f\left(x\right) + g\left(x\right)\right\} = f'\left(x\right) + g'\left(x\right)

Therefore:

\dfrac{d}{dx}\left\{ 2\cdot 6^x + \cos{\left(x\right)} \right\} = 2\cdot 6^x \cdot \ln{6} -\sin{\left(x\right)}

We have:

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

We also have:

\dfrac{d}{dx}\left\{ 9^x \right\} = 9^x \ln{9}

For a linear combination of differentiable functions:

\dfrac{d}{dx}\left\{f\left(x\right) + g\left(x\right)\right\} = f'\left(x\right) + g'\left(x\right)

Therefore:

\dfrac{d}{dx}\left\{ -3\ln{\left(x\right)} + 9^x \right\} = \dfrac{-3}{x} + 9^x \ln{9}