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Sagot :
Ok first we can split it in two : [tex]e^{x^2+2x}[/tex] and [tex]3x[/tex].
The derivative of [tex]3x[/tex] is 3.
For the first part, we use the chain rule : [tex][f(g(x))]'=g'(x)f'(g(x))[/tex] hence [tex](e^{x^2+2x})'=(x^2+2x)'e^{x^2+2x}[/tex] (since the derivative of the exponential is itself) hence [tex]g'(x)=(2x+2)e^{x^2+2x}+3[/tex]
The derivative of [tex]3x[/tex] is 3.
For the first part, we use the chain rule : [tex][f(g(x))]'=g'(x)f'(g(x))[/tex] hence [tex](e^{x^2+2x})'=(x^2+2x)'e^{x^2+2x}[/tex] (since the derivative of the exponential is itself) hence [tex]g'(x)=(2x+2)e^{x^2+2x}+3[/tex]
[tex]g(x)=e^{x^2+2x}+3x\\
g'(x)=e^{x^2+2x}\cdot(2x+2)+3\\
g'(x)=2e^{x^2+2x}(x+1)+3
[/tex]
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