Let f(x) =sin(x) /1+x^2. Let y^(n) denote the n^th derivative of f(x) at x=0 then the value of y^(100) +9900y^98) is?

Answer 1

Do not try to differentiate 100 times. If a question seems too hard or to have too many steps, there is usually a trick that you're missing.

The trick here is that the derivative of sin(x) goes in a cycle.

(1) sin'(x)=cos(x)

(2) sin''(x)=cos'(x)= -sin(x)

(3) from 1 and 2, sin'''(x)= -cos(x)

(4) The fourth derivative of sin(x) is the derivative of -cos(x), which we know from 2 equals sin(x)! So we're back where we started! The fourth derivative of sin(x) is sin(x). Therefore, any order derivative of sin(x), if that order is divisible by 4, is sin(x)! We could also say that the even derivatives of sin(0), the second, fourth, sixth, etc derivatives, are 0.

y^(100)=sin⁽¹⁰⁰⁾(0)=sin⁽⁴⁾(0)=0

y^(98)=sin⁽⁹⁸⁾(0)=sin''⁽⁹⁶⁾(0)=sin''(0)=0

y^(100)+9900y^(98)=0+9900*0=0