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$$f(x) = x^2 \sqrt{5 - x^2}$$

Find the derivative at $(1, 2)$.

\begin{align} \frac{d}{dx} \left[ x^2 \sqrt{5 - x^2} \right] & = \frac{d}{dx} \left[ x^2 (5 - x^2)^{1/2} \right] \\ & = x^2 \frac12 (5 - x^2)^{-1/2}(-2x) + (5 - x^2)^{1/2}(2x) \end{align}

The equation is formed using the product rule and the chain rule. The author explained that $(-2x)$ was inserted as "multiply as the inside derivative", and $(2x)$ as the derivative of the first term.

I assume the following: $(-2x)$ is the derivative of the inside derivative which is $(5-x^2)$. $(2x)$ is the derivative of $x^2$.

My question is, why isn't $(-2x)$ multiplied for the 2nd part of the equation (after the $+$ sign)?

Bladewood
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ilovetolearn
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  • The product rule being $\frac{d}{dx} f(x)g(x) = f'(x)g(x) + f(x)g'(x)$ means you do not have to apply the chain rule in the second term, as the entirety of $\sqrt{5-x^2}$ is being left undisturbed. – Theo Diamantakis May 07 '18 at 15:09

6 Answers6

7

Because that's the part where the external $x^2$ is differentiated

$$(fg)'=f'g+fg'$$

Here $f=x^2$ and $g=\sqrt{5-x^2}$, thus $f'=2x$ and $g'=-2x(5-x^2)^{-1/2}$, thus the result, the second part of the equation is $f'g$

user438666
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5

The $-2x$ is the derivative of the argument in the radical, $5-x^2$.

The $2x$ is the derivative of the left term outside the radical, $x^2$. For the second term in the derivative you take the right multiplicand as-is.

John
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4

Lets take a step back and think about what product rule states: $$\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$$

the $-2x$ is the result of applying chain rule to the square root function in your problem. For simplicity, say $$f(x) = x^2, g(x) = \sqrt{5-x^2}$$ Since the term on the LHS of your derivative includes $x^2$, or $f(x)$, we can conclude that the LHS is the $f(x)g'(x)$ term of the derivative.

Mike H
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2

Let $p(x)=x^2$, $q(x)=\sqrt{x}$ and $r(x)=5-x^2$. Then $(p(x)\cdot q(r(x))^\prime = (q(r(x)))^\prime\cdot p(x) + q(r(x)) \cdot p^\prime(x) = q^\prime(r(x))\cdot r^\prime(x)\cdot p(x) + q(r(x)) \cdot p^\prime(x)$

Kirill Losev
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The derivative actually reads $f'(x) = u'(x)v(x) + u(x)v'(x)$ with $u(x)=x^2$ and $v(x)=\sqrt{5-x^2}$.

For $u(x)$, we have $u'(x)=2x$.

For $v(x)$, we can rewrite $v(x)=f(g(x))$ where $f(t)=\sqrt{t}$ and $g(x)=5-x^2$, so that $v'(x)=g'(x)f'(g(x))=-\frac{2x}{2\sqrt{5-x^2}}$.

Hence $f'(x)=2x\sqrt{5-x^2}-(x^2)(-\frac{2x}{2\sqrt{5-x^2}})=2x\sqrt{5-x^2}+x^2\frac{2x}{2\sqrt{5-x^2}}$, which is the equation shown.

Ernest F
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There are two different $x^2$ terms in the expression being differentiated. Let’s call the first one $g(x)$ and the second $h(x)$. Then we have $$\begin{align} \left(g(x)\sqrt{5-h(x)}\right)' &= g(x)\left(\sqrt{5-h(x)}\right)' + g'(x)\sqrt{5-h(x)} \\ &= \frac12 g(x)\left(5-h(x)\right)^{-1/2}(-h'(x)) + g'(x)\sqrt{5-h(x)}.\end{align}$$ Comparing this to the expression you’re asking about, you can see that the $-2x$ in the first term comes from differentiating the $x^2$ that’s under the radical, while the $2x$ in the second term comes from the $x^2$ outside the radical.

amd
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