How Do You Calculate the Third Derivative of \( y = \frac{1}{x+1} \)?

  • Thread starter Thread starter mo7tn
  • Start date Start date
mo7tn
Messages
1
Reaction score
0

Homework Statement


d3y/dx3 OF Y= 1/X+1


Homework Equations



Y= 1/X+1

The Attempt at a Solution



i TRIED USING THE QUOTIENT RULE BUT BECAME STUCK AFTER DOING- (X+1)D/DX(1) + (1)D/DX(X+1)= (X+1)(1)+(1)(1)=1+1+1= 3 BUT HOW DO GET TO THE THRID DEGREE IM LOST
 
Physics news on Phys.org
y=\frac{1}{x+1}

\frac{dy}{dx}=\frac{1(x+1)-1(1)}{(x+1)^2}

\frac{dy}{dx}==\frac{x}{(x+1)^2}
then just differentiate that twice again and you'll get \frac{d^3y}{dx^3}
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top