Question

Given that, y^{sinx} = x^{cos^{2}x}, find:frac{dx}{dy}

Differential equations
ANSWERED
asked 2020-12-01
Given that, \(\displaystyle{y}^{{{\sin{{x}}}}}={x}^{{{{\cos}^{{{2}}}{x}}}},{f}\in{d}:{\frac{{{\left.{d}{x}\right.}}}{{{\left.{d}{y}\right.}}}}\)

Answers (1)

2020-12-02

Use implicit differentiation
Need to know that: \(\displaystyle{\left(\frac{\partial}{\partial}{x}\right)}{y}={\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={y}′\)
\(\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}{\left({3}{y}+{2}{x}\times{\ln{{\left({y}\right)}}}={y}^{{{4}}}+{x}\right.}\) (expand by chain rule/product rule)
\(3y'+2\times\ln(y)+(\frac{2x}{y})\times y'=4y^{3}y'+1\)
\(3y'+\frac{2xy'}{y}-4y^{3}y'=1-\ln y^{2}\) (collect like terms, logarithm power law)
\(yy'3+\frac{2x}{y}-4y^{3}=1-\ln y^{2}\) (factor)
\(\displaystyle{y}'{\left(\frac{{{3}{y}+{2}{x}-{4}{y}^{{4}}}}{{y}}\right)}={1}-{{\ln{{\left({y}\right)}}}^{{2}}}\) (common divisor for the fraction)
\(\displaystyle{y}'={\frac{{{y}-{y}{\ln{{y}}}^{{{2}}}}}{{{3}{y}+{2}{x}-{4}{y}^{{{4}}}}}}\) Answer
(Try symbolab for computational questions like this)

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