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what is the derivative of $\ln(4)$?

I am trying to find the derivative of this equation: $h(x)=\ln(\frac{x^3\cdot e^x}{4})$

by rules of logs I simplified the $h(x)$ to the following: $h(x)=3\ln(x)+x-\ln(4)$

now when I try to find the derivative of that

$h'(x)=3\frac{\mathrm{d}}{\mathrm{d}x}(\ln(x))+\frac{\mathrm{d}}{\mathrm{d}x}(x)- \frac{\mathrm{d}}{\mathrm{d}x}(\ln(4))$

I get this:

$h'(x)=\frac{3}{x} +1 -\frac{1}{4}$

but my TI-89 tells me that the derivative of $h(x)$ is $h'(x)=\frac{3}{x} +1$

Now, does it mean that $\frac{\mathrm{d}}{\mathrm{d}x}(\ln(4)) =0$? or I did something wrong?

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2 Answers

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No, you haven't done anything wrong. $\ln 4$ is a constant. Hence its derivative is $0$.

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Remember the rule for derivatives of $\ln$: $$\frac {d}{dx} \ln(f(x)) = \frac {\frac {d}{dx}f(x)}{f(x)}$$ In your case $f(x)=4$, thus $$\frac {d}{dx} \ln(4) = \frac {\frac {d}{dx} 4}{4} = \frac {0}{4} = 0$$

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