What are the first three derivatives of #(xcos(x)-sin(x))/(x^2)#? Calculus Basic Differentiation Rules Summary of Differentiation Rules 1 Answer Massimiliano Mar 12, 2015 The answer is: #y''=(-x^3cosx+3x^4sinx+6xcosx-6sinx)/x^4#. This is why: #y'=(((cosx+x*(-sinx)-cosx)x^2-(xcosx-sinx)*2x))/x^4=# #=(-x^3sinx-2x^2cosx+2xsinx)/x^4=# #=(-x^2sinx-2xcosx+2sinx)/x^3# #y''=((-2xsinx-x^2cosx-2cosx-2x(-sinx)+2cosx)x^3-(-x^2sinx-2xcosx+2sinx)*3x^2)/x^6=# #=((-x^2cosx)x^3+3x^4sinx+6x^3cosx-6x^2sinx)/x^6=# #=(-x^3cosx+3x^4sinx+6xcosx-6sinx)/x^4#. Answer link Related questions What is a summary of Differentiation Rules? How do you find the derivative of #(e^(2x) - e^(-2x))/(e^(2x) + e^(-2x))#? How do I find the derivative of #y= x arctan (2x) - (ln (1+4x^2))/4#? How do you find the derivative of #y = s/3 + 5s#? What is the second derivative of #(f * g)(x)# if f and g are functions such that #f'(x)=g(x)#... How do you calculate the derivative for #g(t)= 7/sqrtt#? Can you use a calculator to differentiate #f(x) = 3x^2 + 12#? What is the derivative of #ln(x)+ 3 ln(x) + 5/7x +(2/x)#? How do you find the formula for the derivative of #1/x#? How do you find the derivative for #2x^4-1#? See all questions in Summary of Differentiation Rules Impact of this question 6478 views around the world You can reuse this answer Creative Commons License