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I have the following funtion: f(x) = 3-½√(6-4x)

Now, I would have no idea how to calculate the derivative and the answer booklet doesn't make it any more obvious because according to the answers it should be:

f'(x) = 0-½ . -4 . 1/2√(6-4x) = 1/(√6-4x)

can anyone explain to me what is happening and how they get to this answer?

Mark Fijn
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1 Answers1

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HINT

Use that by chain rule

$$f(x)=c+\sqrt{g(x)}\implies f’(x)=\frac{g’(x)}{2\sqrt{g(x)}}$$

user
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  • So, if this is a standard formula, I can use this on any kind of square root? – Mark Fijn Mar 20 '18 at 18:32
  • Yes the derivative of the constant is zero and the derivative of $\sqrt{g(x)}$ is given by the above expression. Can you derive it by chain rule $f(g(x))=f’(g(x))\cdot g’(x)$? – user Mar 20 '18 at 18:35
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    It is not a standard formula. The standard formula for the chain rule is $f(g(x)) = f'(g(x))\cdot g'x$. This answer shows how it applies to your particular problem, so do not memorize it! It leaves out that $f(x)= c+g(x)^{1/2} \to f'(x) =0 + \frac 12g(x)^{-1/2}\cdot g'(x) $ (by the power rule & chain rule), and this can be written $\frac 12 \cdot \frac 4{\sqrt{g(x)}}$ where $g(x) = 6-4x$. – amWhy Mar 20 '18 at 18:37
  • I'm affraid i don't understand this: f(g(x))=f′(g(x))⋅g′(x) – Mark Fijn Mar 20 '18 at 18:38
  • I really should not have chosen the hardest math i could choose... :) – Mark Fijn Mar 20 '18 at 18:39
  • @amWhy Who said that it is the general formula for chain rule here? Not me of course. I’ve indicated the derivative for $\sqrt{g(x)}$ which is obtained by chain rule. – user Mar 20 '18 at 18:40
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    @gimusi I did, I asked if it was a general formula :) – Mark Fijn Mar 20 '18 at 18:43
  • The thing I don't understand is the g ' (x) part, what are you doing in that part? – Mark Fijn Mar 20 '18 at 18:43
  • @gimusi, I said "this answer shows how it [the chain rule] applies to your particular problem" and meant only that it is not the "general rule" I was asnwering Mark's first comment here. – amWhy Mar 20 '18 at 18:44
  • Mark. Say $g(x) = 6-4x$. What is $g'(x)$? Step by step, $g'(x) = 0+-4 = -4$. – amWhy Mar 20 '18 at 18:46
  • oh wow that was really simple – Mark Fijn Mar 20 '18 at 18:46
  • @MarcFijn it is a general formula for the square root obtained from the more general chain rule formula. Do you know what is the derivative of $\sqrt x$? – user Mar 20 '18 at 18:47
  • And I forgot the negative sign in my result above, which should be $$\frac 12\cdot -\frac 4{\sqrt{6-4x}} = -\frac{2}{\sqrt{6-4x}}$$ – amWhy Mar 20 '18 at 18:48
  • The derivative of squareroot x is x^-1/2 right? – Mark Fijn Mar 20 '18 at 18:48
  • Almost Mark: The derivative of $\sqrt x = x^{1/2}$ is $\frac 12 \cdot x^{-1/2} = \frac 12\cdot \frac 1{x^{1/2}} = \frac{1}{2\sqrt x}$ – amWhy Mar 20 '18 at 18:50
  • @MarkFijn Let use $(x^n)’=nx^{n-1}$ – user Mar 20 '18 at 18:50
  • @amWhy I think I'm starting to understand now – Mark Fijn Mar 20 '18 at 18:52