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.
$\endgroup$ 24 Answers
$\begingroup$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:
$$\frac{d}{dx}(1-2x)^{1/2}=\frac{d}{dx}(1-2x)\cdot=\frac{(1-2x)^{-1/2}}{2}\cdot(-2)$$
$\endgroup$ 8 $\begingroup$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.
$\endgroup$ $\begingroup$$$\frac{d}{dx}(1-2x)^{1/2}=\frac{(1-2x)^{-1/2}}{2^{-1/2}}(-2)$$
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 $\frac{x^{-1/2}}{2^{-1/2}}$.
$\endgroup$ $\begingroup$$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
$\frac{1}{2.\sqrt{1-2x}}.g'(x)=\frac{1}{2\sqrt{1-2x}}.(-2)=\frac{-1}{\sqrt{1-2x}}$
