Text Given y=(12x−5x2)(3+4x2)y=(12x-5x^{2})(3+4x^{2})y=(12x−5x2)(3+4x2), compute y′(x)y'(x)y′(x) and its value at x=1x=1x=1. Solution Product rule with u=12x−5x2u=12x-5x^{2}u=12x−5x2, u′=12−10xu'=12-10xu′=12−10x, v=3+4x2v=3+4x^{2}v=3+4x2, v′=8xv'=8xv′=8x: y′=u′v+uv′=(12−10x)(3+4x2)+(12x−5x2)(8x).y'=u'v+uv'=(12-10x)(3+4x^{2})+(12x-5x^{2})(8x).y′=u′v+uv′=(12−10x)(3+4x2)+(12x−5x2)(8x). Expanding: y′=(36−30x+48x2−40x3)+(96x2−40x3)=36−30x+144x2−80x3.y'=(36-30x+48x^{2}-40x^{3})+(96x^{2}-40x^{3})=36-30x+144x^{2}-80x^{3}.y′=(36−30x+48x2−40x3)+(96x2−40x3)=36−30x+144x2−80x3. At x=1x=1x=1: y′(1)=36−30+144−80=70y'(1)=36-30+144-80=70y′(1)=36−30+144−80=70. y′(x)=−80x3+144x2−30x+36,y′(1)=70\boxed{y'(x)=-80x^{3}+144x^{2}-30x+36,\qquad y'(1)=70}y′(x)=−80x3+144x2−30x+36,y′(1)=70