Find the point on the curve closest to the point (1,3)

Find the point on the curve y=1.04x^2sin(1.05x) which is closest to the point (1,3)

I'm guessing I need to derive something and optimize but I don't even know where to start.

By Pythagoras, the distance from that point to a point on that curve is:
z^2 = x^2 + y^2
z = sqrt(x^2 + y^2)
z = sqrt(x^2 + (1.04x^2sin(1.05x))^2)
z = sqrt(x^2 + 1.0816*x^4*sin^2(1.05x))
Let a = 1.05a, so da/dx = 1.05
Let b = sin(1.05x) = sin(a), so db/da = cos(a) = cos(1.05x)
By the Chain Rule:
db/dx = db/da * da/dx
db/dx = cos(1.05x) * 1.05
db/dx = 1.05*cos(1.05x)
Let c = sin^2(1.05x) = b^2, so dc/db = 2b = 2*sin(1.05x)
By the Chain Rule:
dc/dx = dc/db * db/dx
dc/dx = 2*sin(1.05x) * 1.05*cos(1.05x)
dc/dx = 2.1*sin(1.05x)*cos(1.05x)
Let d = 1.0816x^4, so d' = 4.3264x^3
Let e = 1.0816*x^4*sin^2(1.05x) = d*c, so we use the Product Rule:
e' = d'c + dc'
e' = (4.3264x^3)(sin^2(1.05x)) + (1.0816x^4)(2.1*sin(1.05x)*cos(1.05x))
Let f = x^2 + 1.0816*x^4*sin^2(1.05x) = x^2 + e, therefore:
df/dx = 2x + e'
df/dx = 2x + (4.3264x^3)(sin^2(1.05x)) + (1.0816x^4)(2.1*sin(1.05x)*cos(1.05x))
So z = sqrt(f) = f^0.5, so therefore:
dz/df = 0.5f^(-0.5)
dz/df = 1 / (2 * sqrt(x^2 + 1.0816*x^4*sin^2(1.05x)))
By the Chain Rule:
dz/dx = dz/df * df/dx
dz/dx = (1 / (2 * sqrt(x^2 + 1.0816*x^4*sin^2(1.05x)))) * (2x + (4.3264x^3)(sin^2(1.05x)) + (1.0816x^4)(2.1*sin(1.05x)*cos(1.05x)))
Now set that to zero and solve for x.

Yes, an awkward problem: it was evident that the easiest way to find a solution was through numerical approximation methods, and the difficulty there was to keep the calculations reasonably simple while achieving a controlled convergence towards the final result. The method I used involved the calculation of the gradient of the curve

f(x) = 1.04*x^2*sin(1.05*x)

from the coordinates of the fixed point P at (1, 3) and a trial point A on f(x) at [x, f(x)]. Since AP is normal to f(x) when A is at the minimum distance from P, its slope may be compared with that obtained from the differentiation of f(x) at the point A.to determine when that condition is satisfied. The gradient at A is given by