Differentiation_II(7.4-7.5)

Find derivatives by implicit differentiation

  • Theory
  • Examples
    Differentiate \(y = {(3{x^2} - 1)^2}{({x^4} + 4)^3}{(5x + 6)^5}\) with respect to x.
  • Solutions
    \(\begin{array}{1}y = {(3{x^2} - 1)^2}{({x^4} + 4)^3}{(5x + 6)^5}\\\
    ln y = 2\ln (3{x^2} - 1) + 3\ln ({x^4} + 4) + 5\ln {(5x + 6)^{^{^{^{}}}}}
    \\\frac{d}{{dx}}(\ln y) = \frac{2}{{3{x^2} - 1}} \cdot (6x) + \frac{3}{{{x^4} + 4}} \cdot (4{x^3}) + \frac{5}{{5x + 6}} \cdot {5^{^{^{^{^{^{}}}}}}}\\\frac{1}{y} \cdot \frac{{dy}}{{dx}} = \frac{{12x}}{{3{x^2} - 1}} + \frac{{12{x^3}}}{{{x^4} + 4}} + {\frac{{25}}{{5x + 6}}^{^{^{^{}}}}}\\\frac{{dy}}{{dx}} = (\frac{{12x}}{{3{x^2} - 1}} + \frac{{12{x^3}}}{{{x^4} + 4}} + \frac{{25}}{{5x + 6}}){y^{^{^{^{^{^{}}}}}}}\\ = {\underline{\underline {(\frac{{12x}}{{3{x^2} - 1}} + \frac{{12{x^3}}}{{{x^4} + 4}} + \frac{{25}}{{5x + 6}}){{(3{x^2} - 1)}^2}{{({x^4} + 4)}^3}{{(5x + 6)}^5}}} ^{^{^{}}}}\end{array}\)

Find the second derivative of an explicit function

  • Theory
  • Examples
    (a) If \({x^2}y = \sin x - \cos x\), show that \({x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + y({x^2} + 2) = 0\).
    (b) Find \(\frac{{{d^2}y}}{{d{x^2}}}\).
  • Solutions
    (a)\begin{array}{1}{x^2}y = \sin x - \cos x
    \\{x^2}\frac{{dy}}{{dx}} + 2xy = \cos x + \sin {x^{^{^{}}}}\\{x^2}\frac{{{d^2}y}}{{d{x^2}}} + 2x\frac{{dy}}{{dx}} + 2x\frac{{dy}}{{dx}} + 2y = - \sin x + \cos x\\{x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + 2y = - {x^2}y\\{x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + y({x^2} + 2) = 0\end{array}

    (b)     From (a),
    \(\begin{array}{1}{x^2}\frac{{dy}}{{dx}} + 2xy = \cos x + \sin x\\\frac{{dy}}{{dx}} = {\frac{{\sin x + \cos x - 2xy}}{{{x^2}}}^{}}\end{array}\)
    \(\begin{array}{c}{x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x\frac{{dy}}{{dx}} + y({x^2} + 2) = 0\\{x^2}\frac{{{d^2}y}}{{d{x^2}}} + 4x(\frac{{\sin x + \cos x - 2xy}}{{{x^2}}}) + y({x^2} + 2) = {0^{^{^{^{^{}}}}}}\end{array}\)
    \(\begin{array}{c}\frac{{{d^2}y}}{{d{x^2}}} = \frac{{4(2xy - \sin x - \cos x)}}{{{x^3}}} - \frac{{y({x^2} + 2)}}{{{x^2}}}\\ = {\underline{\underline {\frac{{6xy - 4\sin x - 4\cos x - {x^3}y}}{{{x^3}}}}} ^{}}\end{array}\)