ddx(tan−15x+13−x−6x2)Applyingchainruleddxf(u)=dfdu⋅dudxLetu=5x+13−x−6x2=ddutan−1u⋅ddx(5x+13−x−6x2)Weknowthatddutan−1u=1u2+1Step1:ddutan−1u=1u2+1Step2:ddx(5x+13−x−6x2)=(5x+1)(3−x−6x2)−ddx(3−x−6x2)(5x+1)(3−x−6x2)2=5(3−x−6x2)−(−12x−1)(5x+1)(3−x−6x2)2=30x2+12x+16(3−x−6x2)2Thereforeddx(5x+13−x−6x2)=30x2+12x+16(3−x−6x2)2(Usingquotientruleofdifferentiation)Combiningstep1andstep2:ddutan−1u=1u2+1andddx(5x+13−x−6x2)=30x2+12x+16(3−x−6x2)2=1[5x+13−x−6x2]2+1⋅30x2+12x+16(3−x−6x2)2=15x2+6x+818x4+6x3+5x2+2x+5Henceddx(tan−15x+13−x−6x2)=15x2+6x+818x4+6x3+5x2+2x+5