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$$ \frac{d}{dx} \ln(x+ \sqrt[]{ x^{2} + y^{2} }) $$

What I've done so far:

$$1+\frac{0.5(x^{2})^{-0.5}2x}{x+\sqrt{x^{2}+y^{2}}}$$

$$1+\frac{\frac{x}{(x^{2})^{0.5}}}{x+\sqrt{x^{2}+y^{2}}}$$

Apparently (not my answer) this is supposed to simplify to this if it were correct:

$$\frac{1}{\sqrt{x^{2}+y^{2}}}$$

Could someone show me the next step so I can proceed? Thanks.

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  • $\begingroup$ Is $y$ a constant? $\endgroup$ – beep-boop Sep 19 '14 at 14:10
  • $\begingroup$ yes sorry that was a mistake I've edited it. $\endgroup$ – Osama Qarem Sep 19 '14 at 14:16
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Note that $$\frac d{dx}\ln x =\frac 1x$$

If $z$ is a function of $x$ the chain rule tells us that $$\frac d{dx}\ln z= \frac d{dz} \ln z\cdot \frac {dz}{dx}=\frac 1z\cdot \frac{dz}{dx}$$

This is a useful form for the logarithmic derivative of a function, sometimes written $(\ln z)'=\frac {z'}z$,which is worth noting in its own right for further use.

Now apply this with $z=x+\sqrt{x^2+y^2}$ so that $$\frac {dz}{dx}=1+2x\cdot \frac 12\cdot \frac 1{\sqrt{x^2+y^2}}=\frac{\sqrt{x^2+y^2}+x}{\sqrt{x^2+y^2}}=\frac z{\sqrt{x^2+y^2}}$$

Now it is obvious that $$\frac d{dx}\ln z=\frac 1z\cdot \frac{dz}{dx}=\frac 1{\sqrt{x^2+y^2}}$$

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  • $\begingroup$ This is the nicest, most elegant solution imo. +1 $\endgroup$ – Timbuc Sep 19 '14 at 14:27
  • $\begingroup$ Thanks for answering. I've been trying to understand, how does $1+2x\cdot \frac 12\cdot \frac 1{\sqrt{x^2+y^2}}$ become $\frac{\sqrt{x^2+y^2}+x}{\sqrt{x^2+y^2}}$ shouldn't it simplify to $1+\frac{x}{\sqrt{x^{2}+y^{2}}}$ ? $\endgroup$ – Osama Qarem Sep 19 '14 at 14:41
  • $\begingroup$ @OsamaKerem yes, but then $1=\frac {\sqrt {x^2+y^2}}{\sqrt {x^2+y^2}}$ so that you can put everything over the common denominator $\sqrt {x^2+y^2}$ $\endgroup$ – Mark Bennet Sep 19 '14 at 14:46
  • $\begingroup$ Ah. Alright thank you. $\endgroup$ – Osama Qarem Sep 19 '14 at 14:52
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Note that by the chain rule $$\frac{d}{{dx}}\ln (x + \sqrt {{x^2} + {y^2}} ) = \frac{{1 + \frac{{x + yy'}}{{\sqrt {{x^2} + {y^2}} }}}}{{x + \sqrt {{x^2} + {y^2}} }} = \frac{1}{{\sqrt {{x^2} + {y^2}} }}\frac{{x + \sqrt {{x^2} + {y^2}} + yy'}}{{x + \sqrt {{x^2} + {y^2}} }} $$Now, if $y$ is independent from $x$ you get your result ;)

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Apply the Chain Rule:

$$\left(\log\left(x+\sqrt{x^2+y^2}\right)\right)_x'=\overbrace{\frac1{x+\sqrt{x^2+y^2}}}^{\text{derivative of}\;\log}\cdot\overbrace{\left(1+\frac x{\sqrt{x^2+y^2}}\right)}^{\text{derivative of the argument of}\;\log}=\frac1{\sqrt{x^2+y^2}}$$

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Another Chain rule approach is to use $u=\cosh^{-1} \frac xy\leftrightarrow x=y \cosh u$. Then $x^2+y^2=y^2+y^2\cosh^2 u=y^2 \sinh^2 u$, so $$\ln(x+\sqrt{x^2+y^2})=\ln(y\cosh u+y\sinh u)=\ln(y e^u)=u+\ln y$$ where we have recalled the definitions of hyperbolic sine and cosine. Since $\frac{d}{du}\cosh u =\sinh u$, the chain rule then implies

$$\frac{d}{dx}\ln(x+\sqrt{x^2+y^2})=\frac{d}{du}(u+\ln y)\Big/\frac{dx}{du}= \frac{1}{y \sinh u}=\boxed{ \dfrac{1}{\sqrt{x^2+y^2}}}$$

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