The task is to calculate what we get when we plug $x_{0}=\left( \frac{\pi}{2} \right)$ into the derivative of the function: $$f\left( x \right)=2^{\ln\tan^{2}\left( x-\frac{\pi}{4} \right)}$$ (We also have to calculate the derivative ourselves). When I derived the function I got:$$f'\left( x \right)= \frac{2^{\ln\tan^{2}\left( x-\frac{\pi}{4} \right)+1}\cdot\ln2}{\tan\left( x-\frac{\pi}{4} \right)\cdot\cos^{2}\left(x-\frac{\pi}{4} \right)}$$Upon plugging in $x_0$, I got: $$f'\left( \frac{\pi}{2} \right)=4\cdot\ln2$$This is what the teacher, what multiple people I asked, and what I myself got as the answer. However, when I tried first plugging in $x_0$ into f(x), and then deriving the result, this is what I got: $$f\left( \frac{\pi}{2} \right)=1$$ And as you know, 1 derived is equal to zero. Why don't the solutions match?
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6Did you notice that, if that argument worked, then $f'(x_0)=0$ for every function $f$ and any point of its domain? – José Carlos Santos Sep 10 '23 at 14:29
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Definition of the derivative at $~x_0$: $$\lim_{h \to 0} \frac{f(x_0+h) - f(x_0)}{h}.$$ – user2661923 Sep 10 '23 at 14:37
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Welcome to MSE! <> Not quite a duplicate, but the same basic issue: https://math.stackexchange.com/questions/529050 – Andrew D. Hwang Sep 10 '23 at 14:45
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Thanks for all the help guys! – Anjonimus Sep 10 '23 at 14:50
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When you take the derivative of a function, you take it with respect to a variable - in early calculus classes, this is typically $x$. For this to work, you need the function to depend on that variable. By substituting before differentiating, you are removing the thing that you are trying to differentiate by. This leaves you with a constant function, and the derivative of a constant is always zero, which is clearly not going to be correct.
ConMan
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I see, makes sense. I asked this because I was told that you can first input an x, then derive, or did I mishear something? – Anjonimus Sep 10 '23 at 14:41
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