How do you differentiate #(x^2+3)sqrt(x+2)#?

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Answer 1 · 2018-03-13T15:37:31

#d/dx ( (x^2+3)sqrt(x+2)) =(5x^2 +8x +3)/(2sqrt(x+2))#

#d/dx ( (x^2+3)sqrt(x+2)) = (d/dx(x^2+3)) sqrt(x+2) + (x^2+3)(d/dx sqrt(x+2))#

#d/dx ( (x^2+3)sqrt(x+2)) =2x sqrt(x+2) + (x^2+3)/(2sqrt(x+2))#

Now simplifying:

#d/dx ( (x^2+3)sqrt(x+2)) =(4x (x+2) + (x^2+3))/(2sqrt(x+2))#

#d/dx ( (x^2+3)sqrt(x+2)) =(4x^2 +8x + x^2+3)/(2sqrt(x+2))#

#d/dx ( (x^2+3)sqrt(x+2)) =(5x^2 +8x +3)/(2sqrt(x+2))#