How do you find the second derivative of #y=1/(t^2+1)#?

2 Answers
Anjali G
Apr 8, 2017

#y''=frac{2(t^2-1)}{(t^2+1)^3}#

Explanation:

#y=1/(t^2+1)#

#y=(t^2+1)^(-1)#

Chain rule:
#y'=-(t^2+1)^(-2)(2t)#

#y'=frac{-2t}{(t^2+1)^2}#

Quotient rule and chain rule:
#y'' = frac{(t^2+1)^2(-2)-(-2t)(2)(t^2+1)(2t)}{(t^2+1)^4}#

#y''=frac{color(red)((t^2+1))[-2(t^2+1)+4t^2]}{color(red)((t^2+1)^4)}#

#y''=frac{-2t^2-2+4t^2}{(t^2+1)^3}#

#y''=frac{2t^2-2}{(t^2+1)^3}#

#y''=frac{2(t^2-1)}{(t^2+1)^3}#

Jim G.
Apr 8, 2017

#(d^2y)/(dt^2)=(6t^2-2)/(t^2+1)^3#

Explanation:

One way is to differentiate using the #color(blue)"chain and product rules"#

#"Express " 1/(t^2+1)=(t^2+1)^-1#

differentiate using the #color(blue)"chain rule"#

#"given " y=f(g(x))" then"#

#• dy/dx=f'(g(x))xxg'(x)#

#rArrdy/dt=-(t^2+1)^-2xxd/dt(t^2+1)#

#color(white)(rArrdy/dt)=-2t(t^2+1)^-2#

#"differentiate "dy/dt" using the " color(blue)"product rule"#

#"given " f(x)=g(x).h(x)" then"#

#• f'(x)=g(x)h'(x)+h(x)g'(x)#

#g(t)=-2trArrg'(t)=-2#

#h(t)=(t^2+1)^-2rArrh'(t)=-2(t^2+1)^-3(2t)#

#color(white)(XXXXXXXXXXXXXX)=-4t(t^2+1)^-3#

#rArr(d^2y)/(dt^2)=-2t.-4t(t^2+1)^-3+(t^2+1)^-2(-2)#

#color(white)(rArrd^2y/dt^2)=8t^2(t^2+1)^-3-2(t^2+1)^-2#

#color(white)(rArrd^2y/dt^2)=2(t^2+1)^-3(4t^2-t^2-1)#

#color(white)(rArrd^2y/dt^2)=(6t^2-2)/(t^2+1)^3#

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