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)
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)
【General guidance】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