Please help!!!!!!
2006-11-07 05:46:25 · 6 answers · asked by wat~ 3 in Science & Mathematics ➔ Mathematics
According to the chain rule and to the rule for the derivative of a product, we have y' = (x^2 +3) e^(x^2+1) (2x) + (2x) e^(x^2 +1) = e^(x^2 +1) (2x^3 + 6x + 2x) = e^(x^2 +1) (2x^3 + 8x)
2006-11-07 05:56:58 · answer #1 · answered by Steiner 7 · 0⤊ 0⤋
y is the product of two functions of x, f(x)=(x^2 + 3) and g(x)=e^(x^2 + 1). Using the product rule, y' = f(x) * g'(x) + f'(x)* g(x).
f ' (x) = 2x and g'(x) = [2x*e^(x^2 + 1)]. To find g(x), the chain rule had to be used because it was a function within a function.
2006-11-07 13:55:13 · answer #2 · answered by arcadiaz04 2 · 0⤊ 0⤋
Use product rule:
dy/dx =(x^2+3)(2xe^(x^2+1))+2xe^(x^2+1)
=[2xe^(x^2+1)](x^2 + 4)
2006-11-07 13:54:41 · answer #3 · answered by ironduke8159 7 · 0⤊ 0⤋
y' = 2xe^(x^2+1) + 2x(x^2+3)e^(x^2+1)
2006-11-07 14:01:48 · answer #4 · answered by ag_iitkgp 7 · 0⤊ 0⤋
2x(x^2+4)e^(x^2+1) is the answer
apply u-v rule
the fuction has two terms x^2+3 and e^(x^2+1)
dy/dx=Vdu/dx+udv/dx
=2x*e^(x^2+1)+(x^2+3)e^(x^2+1)
=2x(x^2+4)e^(x^2+1)
2006-11-07 13:56:51 · answer #5 · answered by Krishna D 2 · 0⤊ 0⤋
(2x)*(x^2 + 4)*e^(x^2 + 1)
2006-11-07 13:50:10 · answer #6 · answered by Anonymous · 0⤊ 0⤋