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Differentiate respect to $x$
$x+\sqrt{1-x^2}$

I did it,
$(1-x^2)^\frac{1}{2}$
$\frac{1}{2}(1-2x)$

but the answer is $1-\frac{x}{\sqrt{1-x^2}}$

thanks

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You know how to use the chain rule, I suppose. –  J. M. Apr 29 '12 at 10:58
    
yea. $u=(1-x^2)^\frac{1}{2}$ and $y=u^\frac{1}{2}$ , it is right? –  Sb Sangpi Apr 29 '12 at 11:03
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4 Answers 4

up vote 1 down vote accepted

If you have a function f(x):

$$f(x) = g(x)^m$$

then the derivative of f(x) with respect to x is:

$${df \over dx}(x) = m \times g(x)^{(m-1)} \times {dg \over dx}(x)$$

where ${dg \over dx}(x)$ is the derivative of g(x) with respect to x.

So, if you have:

$$f(x) = (1-x^2)^{1 \over 2}$$

then the derivative of f(x) with respect to x is:

$${df \over dx}(x) = {1 \over 2} \times {(1-x^2)}^{-{1 \over 2}} \times -2x = - {x \over \sqrt {1-x^2}}$$

More generally, chain rule works when you have $f(x) = h(g(x))$ then ${df \over dx}(x) = {dh \over dx}(g(x)) \times {dg \over dx}(x)$.

In your particular example, $h(x) = x^{1 \over 2}$ and $g(x) = 1-x^2$. So ${dh \over dx}(x) = {1 \over 2} \times x^{-{1 \over 2}}$ and ${dg \over dx}(x) = -2x$.

Therefore putting it all together gives ${df \over dx}(x) = {1 \over 2} \times (1-x^2)^{-{1 \over 2}} \times -2x$

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The function $\sqrt{1-x^2}$ actually is a composite function consisting of: $$f(x)=\sqrt{x} \ and \ g(x)=1-x^2 \ with \ f'(x)=\frac{0.5}{\sqrt{x}}, g'(x)=-2x$$ Your function is $f(g(x))$ and therefore its derivative according to the chain rule is: $$\frac{d(f(g(x))}{dx}=f'(g(x))\cdot g'(x)=\frac{0.5}{\sqrt{1-x^2}}\cdot(-2x)=\frac{-x}{\sqrt{1-x^2}}$$

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If $y=\sqrt{1-x^2}$, then $y$ is the upper half of the unit circle $x^2+y^2=1$. An arbitrary point $(x,y)$ on this circle has radial slope $\frac{y}{x}$, and the tangent slope is the negative reciprocal; that is, $$ y'=\frac{dy}{dx}=-\left(\frac{y}{x}\right)^{-1}=-\frac{x}{y} $$ without even needing to use the rules of calculus! If we want to use those rules, then we do instead $$ \eqalign{y' &=\frac{d}{dx}\left(1-x^2\right)^{1/2}\\ &=\frac12\left(1-x^2\right)^{-1/2}\frac{d}{dx}\left(1-x^2\right) \qquad\text{power & chain rules}\\ &=\frac12\cdot\frac{-2x}{\left(1-x^2\right)^{1/2}}\\ &=\frac12\cdot\frac{-2x}{\left(1-x^2\right)^{1/2}}\\ &=\frac{x}{y}\\ } $$ Finally, the answer to your problem is then $$ \frac{d}{dx}\left(x+y\right) =\frac{dx}{dx}+\frac{dy}{dx} =1+y' =1-\frac{x}{y} $$ or, equivalently, $$ 1-\frac{x}{y} =1-\frac{x}{\left(1-x^2\right)^{1/2}} =\frac{\left(1-x^2\right)^{1/2}-x}{\left(1-x^2\right)^{1/2}} =1-x\left(1-x^2\right)^{-1/2} \,. $$

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To differentiate $\sqrt{1-x^2}$

let $g(x)= 1-x^2$

now we have to differentiate $(gx)^{\frac{1}{2}}$

$\frac{d}{dx} (gx)^{\frac{1}{2}}=\frac{1}{2}(gx)^{\frac{-1}{2}}.g'(x)$

plug in the terms to get $-\frac{x}{\sqrt{1-x^2}}$

hence,

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

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