Question

39-42 Find \(y''\) by implicit differentiation/ Simplify where possible.

41. \({\bf{sin}}\,y + {\bf{cos}}x = {\bf{1}}\)

Short answer

The value of \(y''\) is \(\frac{{{{\cos }^2}y\cos x + \sin y{{\sin }^2}x}}{{{{\left( {\cos y} \right)}^3}}}\).

Step-by-step solution

Differentiate the given equation with respect to x

Differentiate the equation \(\sin y + \cos x = 1\) on both sides.

\(\begin{aligned}\cos y\left( {\frac{{{\rm{d}}y}}{{{\rm{d}}x}}} \right) - \sin x &= 0\\\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= \frac{{\sin x}}{{\cos y}}\end{aligned}\)

Find the value of \(y''\)

Find the value of \(y''\) or \(\frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}}\) by differentiating the equation \(\frac{{{\rm{d}}y}}{{{\rm{d}}x}} = \frac{{\sin x}}{{\cos y}}\).

\(\begin{aligned}\frac{{{{\rm{d}}^2}y}}{{{\rm{d}}{x^2}}} &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{\sin x}}{{\cos y}}} \right)\\ &= \frac{{\cos y\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\sin x} \right) - \sin x\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\cos y} \right)}}{{{{\left( {\cos y} \right)}^2}}}\\ &= \frac{{\cos y\cos x + \sin x\sin y\frac{{{\rm{d}}y}}{{{\rm{d}}x}}}}{{{{\left( {\cos y} \right)}^2}}}\\ &= \frac{{\cos y\cos x + \sin x\sin y\left( {\frac{{\sin x}}{{\cos y}}} \right)}}{{{{\left( {\cos y} \right)}^2}}}\\ &= \frac{{{{\cos }^2}y\cos x + \sin y{{\sin }^2}x}}{{{{\left( {\cos y} \right)}^3}}}\end{aligned}\)

So, the value of \(y''\) is \(\frac{{{{\cos }^2}y\cos x + \sin y{{\sin }^2}x}}{{{{\left( {\cos y} \right)}^3}}}\).