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Derivative of $$$x^{4} \cos{\left(x \right)}$$$
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Your Input
Find $$$\frac{d}{dx} \left(x^{4} \cos{\left(x \right)}\right)$$$.
Solution
Let $$$H{\left(x \right)} = x^{4} \cos{\left(x \right)}$$$.
Take the logarithm of both sides: $$$\ln\left(H{\left(x \right)}\right) = \ln\left(x^{4} \cos{\left(x \right)}\right)$$$.
Rewrite the RHS using the properties of logarithms: $$$\ln\left(H{\left(x \right)}\right) = 4 \ln\left(x\right) + \ln\left(\cos{\left(x \right)}\right)$$$.
Differentiate separately both sides of the equation: $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{d}{dx} \left(4 \ln\left(x\right) + \ln\left(\cos{\left(x \right)}\right)\right)$$$.
Differentiate the LHS of the equation.
The function $$$\ln\left(H{\left(x \right)}\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = H{\left(x \right)}$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(H{\left(x \right)}\right)\right)}$$The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(H{\left(x \right)}\right) = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(H{\left(x \right)}\right)$$Return to the old variable:
$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(u\right)}} = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{{\color{red}\left(H{\left(x \right)}\right)}}$$Thus, $$$\frac{d}{dx} \left(\ln\left(H{\left(x \right)}\right)\right) = \frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}}$$$.
Differentiate the RHS of the equation.
The derivative of a sum/difference is the sum/difference of derivatives:
$${\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x\right) + \ln\left(\cos{\left(x \right)}\right)\right)\right)} = {\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x\right)\right) + \frac{d}{dx} \left(\ln\left(\cos{\left(x \right)}\right)\right)\right)}$$Apply the constant multiple rule $$$\frac{d}{dx} \left(c f{\left(x \right)}\right) = c \frac{d}{dx} \left(f{\left(x \right)}\right)$$$ with $$$c = 4$$$ and $$$f{\left(x \right)} = \ln\left(x\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(4 \ln\left(x\right)\right)\right)} + \frac{d}{dx} \left(\ln\left(\cos{\left(x \right)}\right)\right) = {\color{red}\left(4 \frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \frac{d}{dx} \left(\ln\left(\cos{\left(x \right)}\right)\right)$$The derivative of the natural logarithm is $$$\frac{d}{dx} \left(\ln\left(x\right)\right) = \frac{1}{x}$$$:
$$4 {\color{red}\left(\frac{d}{dx} \left(\ln\left(x\right)\right)\right)} + \frac{d}{dx} \left(\ln\left(\cos{\left(x \right)}\right)\right) = 4 {\color{red}\left(\frac{1}{x}\right)} + \frac{d}{dx} \left(\ln\left(\cos{\left(x \right)}\right)\right)$$The function $$$\ln\left(\cos{\left(x \right)}\right)$$$ is the composition $$$f{\left(g{\left(x \right)} \right)}$$$ of two functions $$$f{\left(u \right)} = \ln\left(u\right)$$$ and $$$g{\left(x \right)} = \cos{\left(x \right)}$$$.
Apply the chain rule $$$\frac{d}{dx} \left(f{\left(g{\left(x \right)} \right)}\right) = \frac{d}{du} \left(f{\left(u \right)}\right) \frac{d}{dx} \left(g{\left(x \right)}\right)$$$:
$${\color{red}\left(\frac{d}{dx} \left(\ln\left(\cos{\left(x \right)}\right)\right)\right)} + \frac{4}{x} = {\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right) \frac{d}{dx} \left(\cos{\left(x \right)}\right)\right)} + \frac{4}{x}$$The derivative of the natural logarithm is $$$\frac{d}{du} \left(\ln\left(u\right)\right) = \frac{1}{u}$$$:
$${\color{red}\left(\frac{d}{du} \left(\ln\left(u\right)\right)\right)} \frac{d}{dx} \left(\cos{\left(x \right)}\right) + \frac{4}{x} = {\color{red}\left(\frac{1}{u}\right)} \frac{d}{dx} \left(\cos{\left(x \right)}\right) + \frac{4}{x}$$Return to the old variable:
$$\frac{\frac{d}{dx} \left(\cos{\left(x \right)}\right)}{{\color{red}\left(u\right)}} + \frac{4}{x} = \frac{\frac{d}{dx} \left(\cos{\left(x \right)}\right)}{{\color{red}\left(\cos{\left(x \right)}\right)}} + \frac{4}{x}$$The derivative of the cosine is $$$\frac{d}{dx} \left(\cos{\left(x \right)}\right) = - \sin{\left(x \right)}$$$:
$$\frac{{\color{red}\left(\frac{d}{dx} \left(\cos{\left(x \right)}\right)\right)}}{\cos{\left(x \right)}} + \frac{4}{x} = \frac{{\color{red}\left(- \sin{\left(x \right)}\right)}}{\cos{\left(x \right)}} + \frac{4}{x}$$Simplify:
$$- \frac{\sin{\left(x \right)}}{\cos{\left(x \right)}} + \frac{4}{x} = - \tan{\left(x \right)} + \frac{4}{x}$$Thus, $$$\frac{d}{dx} \left(4 \ln\left(x\right) + \ln\left(\cos{\left(x \right)}\right)\right) = - \tan{\left(x \right)} + \frac{4}{x}$$$.
Hence, $$$\frac{\frac{d}{dx} \left(H{\left(x \right)}\right)}{H{\left(x \right)}} = - \tan{\left(x \right)} + \frac{4}{x}$$$.
Therefore, $$$\frac{d}{dx} \left(H{\left(x \right)}\right) = \left(- \tan{\left(x \right)} + \frac{4}{x}\right) H{\left(x \right)} = x^{3} \left(- x \tan{\left(x \right)} + 4\right) \cos{\left(x \right)}$$$.
Answer
$$$\frac{d}{dx} \left(x^{4} \cos{\left(x \right)}\right) = x^{3} \left(- x \tan{\left(x \right)} + 4\right) \cos{\left(x \right)}$$$A