Answer:
B) [tex]2xln(2)(2^{x^2+2})[/tex]
Step-by-step explanation:
[tex]y=2^{x^2+2}[/tex]
[tex]y'=\frac{d}{dx}(2^{x^2+2})[/tex]
[tex]y'=\frac{d}{dx}(e^{(x^2+2)*ln(2)})[/tex]
[tex]y'=\frac{d}{dx}(e^{(x^2ln(2)+2ln(2))})[/tex]
[tex]y'=2xln(2)*e^{(x^2+2)(ln(2)}[/tex]
[tex]y'=2xln(2)(2^{x^2+2})[/tex]
Therefore, [tex]2xln(2)(2^{x^2+2})[/tex] is the slope of the tangent line
