Har denne oppgaven:
La $x = g(y)$ være inversen til $y = f(x)$ ... Da er $ x = g(f(x))$ .. bruk dette til å vise at:
$\frac{dg}{dy}(f(x)) = \frac{1}{\frac{df}{dx}(x)}$, og
$\frac{d^{2}g}{dy^{2}}(f(x)) = -\frac{\frac{d^{2}f}{dx^{2}}(x)}{(\frac{df}{dx}(x))^{3}}$
Så jeg synes notasjonen er litt forvirrende, men dette har jeg prøvd:
Benytter meg av formelen $\frac{d}{dx}f^{-1}(x) = \frac{1}{f'(f^{-1}(x))}$
Da blir: $\frac{dg}{dx}(f(x)) = \frac{1}{\frac{df}{dx}(g(y))} = \frac{1}{\frac{df}{dx}(x)}$
Så vidt jeg forstår skal jeg derivere igjen.. da får jeg vha. kjerneregelen:
$\frac{d}{dx}\left[\frac{1}{\frac{df}{dx}g(y)}\right] = \frac{-\left[\frac{df}{dx}g(y) + y*g(y)'\right]}{\left[\frac{df}{dx}g(y)\right]^{2}} = \frac{-\left[\frac{df}{dx}g(y) + y\left[\frac{1}{\frac{df}{dx}g(y)}\right] \right]}{\left[\frac{df}{dx}g(y)\right]^{2}} = \frac{-\left[\frac{(\frac{df}{dx}g(y))^{2} + y}{\frac{df}{dx}g(y)}\right]}{\left[\frac{df}{dx}g(y)\right]^{2}} = \frac{-\left[\left(\frac{df}{dx}g(y)\right)^{2} + y\right]}{\left[\frac{df}{dx}g(y)\right]^{3}}$
$= \frac{-\left[\left(\frac{df}{dx}(x)\right)^{2} + y\right]}{\left[\frac{df}{dx}(x)\right]^{3}}$
sorry at det ser stygt ut, håper det ikke er for stygt?. Jeg får altså ikke det samme resultatet som dere ser. Hvor går jeg feil?
Deriverte av invers funksjon
Moderators: Vektormannen, espen180, Aleks855, Solar Plexsus, Gustav, Nebuchadnezzar, Janhaa
Notasjonen din gjør det eklere å vise. Du skal vise at
[tex]\frac{\mathrm{d}}{\mathrm{d}y}\left[g(f(x))\right] = \frac{1}{\frac{\mathrm{d}}{\mathrm{d}x}\left[f(x)\right]}[/tex] ?
[tex]\frac{\mathrm{d}}{\mathrm{d}y}\left[g(f(x))\right] =\frac{\mathrm{d}g(f(x))}{\mathrm{d}x}\cdot\frac{\mathrm{d}x}{\mathrm{d}y} = \frac{\mathrm{d}x}{\mathrm{d}x}\cdot\frac{\mathrm{d}x}{\mathrm{d}y} = 1\cdot\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}} = \frac{1}{\frac{\mathrm{d}}{\mathrm{d}x}\left[f(x)\right]}[/tex]
[tex]\frac{\mathrm{d}^2}{\mathrm{d}y^2}\left[g(f(x))\right] = \frac{\mathrm{d}}{\mathrm{d}y}\cdot\frac{\mathrm{d}}{\mathrm{d}y}\left[g(f(x)\right] = \frac{\mathrm{d}x}{\mathrm{d}y}\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}\right] = \frac{\mathrm{d}x}{\mathrm{d}y}\cdot\left[\frac{-\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\right] = \frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}\cdot\left[\frac{-\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\right] =-\frac{\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^3}[/tex]
[tex]\frac{\mathrm{d}}{\mathrm{d}y}\left[g(f(x))\right] = \frac{1}{\frac{\mathrm{d}}{\mathrm{d}x}\left[f(x)\right]}[/tex] ?
[tex]\frac{\mathrm{d}}{\mathrm{d}y}\left[g(f(x))\right] =\frac{\mathrm{d}g(f(x))}{\mathrm{d}x}\cdot\frac{\mathrm{d}x}{\mathrm{d}y} = \frac{\mathrm{d}x}{\mathrm{d}x}\cdot\frac{\mathrm{d}x}{\mathrm{d}y} = 1\cdot\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}} = \frac{1}{\frac{\mathrm{d}}{\mathrm{d}x}\left[f(x)\right]}[/tex]
[tex]\frac{\mathrm{d}^2}{\mathrm{d}y^2}\left[g(f(x))\right] = \frac{\mathrm{d}}{\mathrm{d}y}\cdot\frac{\mathrm{d}}{\mathrm{d}y}\left[g(f(x)\right] = \frac{\mathrm{d}x}{\mathrm{d}y}\frac{\mathrm{d}}{\mathrm{d}x}\left[\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}\right] = \frac{\mathrm{d}x}{\mathrm{d}y}\cdot\left[\frac{-\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\right] = \frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}\cdot\left[\frac{-\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2}\right] =-\frac{\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}{\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^3}[/tex]