How do you find the derivative of #f(x)=(x^2+x)*tan x/x^3#?

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Answer 1 · 2017-05-10T03:33:43

#f'(x)= (x+1)/x^2sec^2x - (x+2)/x^3tanx#

#f(x)=(x^2+x)*tanx/x^3#

#=x(x+1)*tanx/x^3#

#=(x+1)*tanx/x^2#

#= (x+1)/x^2 * tanx#

Applying the product rule:

#f'(x) = (x+1)/x^2* d/dx tanx + tanx * d/dx(x+1)/x^2#

#= (x+1)/x^2*sec^2x + tanx*((x^2*1-(x+1)*2x)/x^4)# Std. differential and quotient rule.

#= (x+1)/x^2*sec^2x + tanx*(x-2x-2)/x^3#

#= (x+1)/x^2*sec^2x - tanx*(x+2)/x^3#