Finding tangent equation of curve y = 3 ^ x at point (1,3)

Finding tangent equation of curve y = 3 ^ x at point (1,3)

y'=ln3*3^x
According to the tangent equation y-y0 = f '(x0) (x-x0)
Y-3 = Ln3 * 3 * (x-1)
So the tangent equation is y = 3 (Ln3) (x-1) + 3

What is the tangent equation of curve y = x ^ 2 + 3 at point (1,4)?
What is y '

Y 'is the derivative of the original equation. Y' = 2x + C, take (1,4) in, and get C = 2, so the tangent equation is y = 2x + 2

The tangent equation of curve y = x ^ 3 + X at point (1,2) is the normal equation

Y = x ^ 3 + X, y '= 3x ^ 2 + 1 let the tangent slope of curve y = x ^ 3 + X at point (1,2) be K
K = y'lx = 1 = 4, the tangent equation is Y-2 = 4 (x-1), that is 4x-y-2 = 0
The normal slope of the curve y = x ^ 3 + X at point (1,2) is - 1 / 4
The normal equation is Y-2 = - 1 / 4 (x-1), that is, x + 4y-9 = 0

Find the tangent equation and normal equation of curve y = 2 / x + X at point (2,3)

y'=-2/x^2 + 1
Substituting x = 2
y'=-2/4+1=1/2
k=1/2
Tangent equation: (Y-3) = 1 / 2 (X-2), that is: 2y-6 = X-2, that is: x-2y + 4 = 0
Normal: K '= - 1 / k = - 1 / (1 / 2) = - 2
Normal equation: (Y-3) = - 2 (X-2), that is: y + 2x-7 = 0