Question

given the curve defined by x^2-siny=y+4
-Find d^2y/dx^2 in terms of x and y only

Answer 1

\[x ^{2}-\sin y=y+4\]
diff. w.r.t. x
\[2x-\cos y \frac{ dy }{dx }=\frac{ dy }{dx }\]
\[\left( 1+\cos y \right)\frac{ dy }{dx }=2x\]
\[\frac{ dy }{ dx }=\frac{ 2x }{ 1+\cos y }\]
diff. w.r.t. x
\[\frac{ d ^{2}y }{ dx ^{2} }=\frac{ \left( 1+\cos y \right)*2-2x \left( -\sin y \right)\frac{ dy }{ dx } }{ \left( 1+\cos y \right)^{2} }\]
\[\frac{ d ^{2}y }{ dx ^{2} }=2\frac{ \left( 1+\cos y \right)+\sin y \frac{ 2x }{ 1+\cos y } }{ \left( 1+\cos y \right)^{2} }\]
\[\frac{ d ^{2}y }{ dx ^{2} }=2\frac{ 1+2\cos y+\cos ^{2}y+2x \sin y }{ \left( 1+\cos y \right)^{3} }\]

Answer 2

How would you find the equation for each tangent line to the curve where y=0? @surjithayer

Answer 3

when y=0
\[x ^{2}-0=0+4,x=2,-2,points are (-2,0) and (2,0)\]
at (-2,0)
slope dy/dx\[slope \frac{ dy }{ dx }=\frac{ 2*-2 }{ 1+1 }=-2\]
eq. of tangent line at (-2,0) is
y-0=-2(x+2)
or y+2x+4=0
similarly we can find tangent line at (2,0)

Answer 4

Would the slope for the tangent line at (2,0) also be -2?

Answer 5

no, it is 2
\[slope at (2,0) \frac{ dy }{dx }=\frac{ 2*2 }{ 1+1 }=2\]