Given that y=2xy = 2^xy=2x show that dydx=2xln2\displaystyle \frac{dy}{dx} = 2^x \ln 2dxdy=2xln2
Find the equation to the tangent of the curve y=3(x2)y = 3^{(x^2)}y=3(x2) at the point (2,81)(2, 81)(2,81)
