What is the derivative of $x + cos (x + y) = 0$ $x + cos(x+y)=0$ ?

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Answer 1 · 2018-07-16T11:10:41

$y'=(1-sin(x+y))/sin(x+y)$

Assuming you mean $y=y(x)$
we get by the chain rule

$1-sin(x+y)-y'sin(x+y)=0$
so we get

$y'=(1-sin(x+y))/sin(x+y)$ if $sin(x+y)ne 0$

Answer 2 · 2018-07-16T11:22:00

$dy/dx=\frac{1}{\sqrt{1-x^2}}-1$

Given equation:

$x+\cos(x+y)=0$

$cos(x+y)=-x$

$x+y=\pi-\cos^{-1}(x)$

$y=\pi-\cos^{-1}(x)-x$

Differentiating above equation w.r.t. $x$ as follows

$\frac{dy}{dx}=\frac{d}{dx}(\pi-\cos^{-1}(x)-x)$

$=0-(-\frac{1}{\sqrt{1-x^2}})-1$

$=\frac{1}{\sqrt{1-x^2}}-1$

What is the derivative of x + cos(x+y)=0x+cos(x+y)=0x + cos(x+y)=0?