# Question Use implicit differentiation to find an equation of the tangent line to the curve at the given point. y sin(16x) = x cos(2y), (𝜋/2, 𝜋/4)

&#12304;General guidance&#12305;The answer provided below has been developed in a clear step by step manner.Step1/2Given, the curve $$\mathrm{{y}{\sin{{\left({16}{x}\right)}}}={x}{\cos{{\left({2}{y}\right)}}}}$$at the point $$\mathrm{{\left(\frac{\pi}{{2}},\frac{\pi}{{4}}\right)}}$$Differentiating both sides with respect to $$\mathrm{{x},}$$we get$$\mathrm{\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({y}{\sin{{\left({16}{x}\right)}}}\right)}=\frac{{d}}{{{\left.{d}{x}\right.}}}{\left({x}{\cos{{\left({2}{y}\right)}}}\right)}}$$Apply the product rule of differentiation,$$\mathrm{{\sin{{\left({16}{x}\right)}}}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}+{16}{y}{\cos{{\left({16}{x}\right)}}}={\cos{{\left({2}{y}\right)}}}-{2}{x}{\sin{{\left({2}{y}\right)}}}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}$$$$\mathrm{{\left({\sin{{\left({16}{x}\right)}}}+{2}{x}{\sin{{\left({2}{y}\right)}}}\right)}\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}={\cos{{\left({2}{y}\right)}}}-{16}{y}{\cos{{\left({16}{x}\right)}}}}$$\( \mathrm{\Rightarrow\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}=\frac{{{\cos{{\left({2}{y}\right)}}}-{16}{y}{\cos{{\left({16}{x}\right)}}}}}{{{\sin{{\left({16}{x} ... See the full answer