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Derivative

This is used for a project and it is pretty late, so I just want to make sure I derived $C_3(x)$ correctly. Thank you!

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    instead of posting this as a question you could as well ask Wolfram Alpha and get an intermediate answer – 2017-02-06
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    Please, use MathJax [http://math.stackexchange.com/help/notation](http://math.stackexchange.com/help/notation) – 2017-02-06
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    @YuriyS I'll try to include better formatting next time – 2017-02-06

3 Answers 3

3

By Chain rule,

$$\dfrac{d\sqrt{a^2+x^2}}{dx}=\dfrac{d\sqrt{a^2+x^2}}{d(a^2+x^2)}\cdot\dfrac{d(a^2+x^2)}{dx}=\dfrac{2x}{2\sqrt{a^2+x^2}}$$

$$\dfrac{d(\sqrt{b^2+(c-x)^2})}{dx}=\dfrac{d(\sqrt{b^2+(c-x)^2})}{b^2+(c-x)^2}\cdot\dfrac{d(b^2+(c-x)^2)}{dx}=?$$

3

Derivative with respect to what? Is the $c_{L}$ a function? You are not giving much of a context.

Assuming $c_{L}$ and $c_{R}$ are constants and $b$ and $c$ are independent of $x$, you are asking for derivative of $\sqrt{a^2+x^2}$ with respect to $x$ and for derivative of $\sqrt{b^2+c^2+x^2-2cx}$ with respect to $x$. In other words, you are asking for derivative of $f(x)^{\frac{1}{2}}$ with respect to $x$.

Then, using standard derivative rules, it is $\frac{1}{2}f(x)^{-\frac{1}{2}}f'(x)$. Hence none of your two derivatives are correct.

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    I should add the derivative and what variables are constants and which are not although from my poor derivative it should be implied that it is with respect to x. Thank you for your answer though – 2017-02-06
2

We have the derivative of $\sqrt{b^2 + (c-x)^2} $ as: $$ \frac {d}{dx}(\sqrt {b^2+(c-x)^2})$$ $$ = \frac {1}{2\sqrt {b^2+(c-x)^2}} \frac {d}{dx}(b^2+(c-x)^2)$$ $$=\frac {1}{2\sqrt {b^2+(c-x)^2}} 2 (c-x) \frac {d}{dx}(c-x) $$ $$=\boxed {-\frac {c-x}{\sqrt {b^2+(c-x)^2}}} $$

Hope it helps.