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I have this differentiation problem, taken from James Stewart's Calculus Early Transcendentals, 7th Ed. Page 205, Exercise 9.

Find the derivative of the below using chain rule.

Given: $$F(x)=\sqrt{1-2x}$$

My solution:

$$\sqrt{1-2x}=(1-2x)^{1/2}$$

$$\frac{d}{dx}(1-2x)^{1/2}=\frac{(1-2x)^{-1/2}}{2^{-1/2}}(-2)$$

Reciprocal of the numerator and denominator will create positive exponents.

$$\frac{2^{1/2}}{(1-2x)^{1/2}}(-2)=\frac{\sqrt{2}}{\sqrt{(1-2x)}}(-2)$$

Multiply by $-2$ for our answer:

$$\frac{-2\sqrt{2}}{\sqrt{(1-2x)}}$$

However the textbook answer is:

$$\frac{-1}{\sqrt{(1-2x)}}$$

Where did I go wrong with my algebra? Thanks for your help.

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    $\begingroup$ I am not entirely sure why you did the division like you did (3rd line) - but you should be multiplying by the power before the derivative. $\endgroup$ – Chinny84 Dec 23 '16 at 15:45
  • $\begingroup$ How did you get the $2^{-\frac12}$ in the denominator? $\endgroup$ – 5xum Dec 23 '16 at 15:45
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Here is the mistake:

$$\frac{d}{dx}(1-2x)^{1/2}=\frac{(1-2x)^{-1/2}}{2}\cdot(-2)$$

Ps:Remmember that $(x^{1/2})'=\frac{1}{2}\cdot x^{-1/2}$

Beeing more specific and using chain rule:

$$(f(g(x)))'=g'(x)\cdot f'(g(x))$$

On your case you can choose $g(x)=1-2x$ and $f(x)=x^{1/2}$ so,

$$g'(x)=-2$$

$$f'(x)=\frac{1}{2}x^{-1/2} \Rightarrow f'(g(x))=\frac{1}{2}(g(x))^{-1/2}=\frac{1}{2}(1-2x)^{-1/2}$$

and then:

$$\frac{d}{dx}(1-2x)^{1/2}=\frac{d}{dx}(1-2x)\cdot=\frac{(1-2x)^{-1/2}}{2}\cdot(-2)$$

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  • $\begingroup$ Thank you. Question: doesn't it hold true that a fraction to a power is equal to the numerator raised to that power divided by the denominator of that same power? $\endgroup$ – baverso Dec 23 '16 at 16:00
  • $\begingroup$ Nope! The general case is $(x^p)'=p\cdot x^{p-1}$ $\endgroup$ – Arnaldo Dec 23 '16 at 16:03
  • $\begingroup$ $\frac{1}{2}(1-2x)^{-1/2}=(\frac{(1-2x)}{2})^(-1/2)=\frac{(1-2x)^{-1/2}}{2^{-1/2}$ $\endgroup$ – baverso Dec 23 '16 at 16:03
  • $\begingroup$ did you get that? $\endgroup$ – Arnaldo Dec 23 '16 at 16:03
  • $\begingroup$ Tried to insert my comment with mathjax but for some reason it's not working. I completely agree with you on the above. My question is if we multiply the term by (1/2) we have that term over 2, raised to the -(1/2) power. Then a property of exponents says that the numerator and denominator each can take on that power (-1/2). $\endgroup$ – baverso Dec 23 '16 at 16:06
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The denominator of $2^{-\frac{1}{2}}$ is incorrect.

The derivative of $x^\frac{1}{2}$ is $\frac{1}{2}x^{-\frac{1}{2}}$, so it should be mulitplied by the constant $\frac{1}{2}$ instead.

Now just substitute $1 - 2x$ into $x$ above and multiply throughout with the derivative of $1 - 2x$ to effect the chain rule.

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$$\frac{d}{dx}(1-2x)^{1/2}=\frac{(1-2x)^{-1/2}}{2^{-1/2}}(-2)$$

is wrong; the factor $2^{-1/2}$ in the denominator should just be $2$ instead - the derivative of $x^{1/2}$ is $\frac{1}{2}x^{-1/2}$, not $\frac{x^{-1/2}}{2^{-1/2}}$.

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$f(g(x))=\sqrt{1-2x} $ and $ g(x)=1-2x$

so now applying chain rule( derivative of $f(g(x))= f'(g(x)).g'(x)$), we get the following
$\frac{1}{2.\sqrt{1-2x}}.g'(x)=\frac{1}{2\sqrt{1-2x}}.(-2)=\frac{-1}{\sqrt{1-2x}}$

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