Step-by-step explanation:
so
\(f(x)=9x^1^0tan^-^1(x)\)
to find \(f'(x)=\)
\(d/dx(9x^1^0tan^-^1(x))\)
differentiation is linear so we solve differentiate separately and pull out constant factors
\([a.u(x)+b.v(x)]'=a.u'(x)+b.v'(x)\)
\(=9.d/dx(x^1^0tan^-^1(x))\)
applying product rule
\([u(x).v(x)]'=u'(x).v(x)+u(x).v'(x)\)
\(=9(d/dx[x^1^0].tan(x)+x^1^0.d/dx[tan(x)])\)
\(=9(10x^9tan(x)+x^1^0.(1/x^2+1))\)
\(=9(10x^9tan(x)+x^1^0/x^2+1)\)
simplify
\(f'(x)=90x^9tan(x)+(9x^1^0/x^2+1)\)
