Implicit differentiation (1 Viewer)

JustRandomThings · Jan 5, 2017

Hey everyone thanks in advance!

Question is attached below

InteGrand · Jan 5, 2017

JustRandomThings said:
Hey everyone thanks in advance!

Question is attached belowView attachment 33689

$\noindent Using implicit differentiation / chain rule, differentiating both sides wrt $x$, we have$

$\begin{align*}y + xy' + \frac{1}{1+\left(x+y\right)^{2}}\cdot \left(1+y'\right) &= 0.\end{align*}$

$\noindent Note that at $x=0$, we have $y=1$, so at this point (substituting $x =0,y=1$ into the previous equation), we have$

$\begin{align*}1+0 + \frac{1}{1+1^{2}} \times \left(1+y'\right) &= 0\\ \Rightarrow y' &= -3.\end{align*}$

$\noindent So the answer is $-3$.$

Implicit differentiation (1 Viewer)

JustRandomThings

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InteGrand

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