How do you find the derivative of #h(x)=1/2(x^2+1)(5-2x)#?

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Answer 1 · 2018-05-10T20:44:44

#h'(x)=x*(5-2x)-(x^2+2)=5x-2x^2-x^2-2=-3x^2+5x-2#

show below

#h(x)=1/2(x^2+1)(5-2x)#

now we will dirvite it

#h'(x)=1/2*[(2x+0)*(5-2x)+(x^2+2)(0-2)]#

#h'(x)=1/2*[2x*(5-2x)+(x^2+2)(-2)]#

#h'(x)=x*(5-2x)-(x^2+2)=5x-2x^2-x^2-2=-3x^2+5x-2#

note that

#y=g(x)*f(x)#

#y'=g(x)*f'(x)+g'(x)*f(x)#