Find the gradient of the tangent to the curve at the point indicated :
Problem 1 :
y = x² + 3x (2, 10)
Solution :
y = x² + 3x
dy/dx = 2x + 3
Put x = 2,
dy/dx = 2(2) + 3
= 4 + 3
dy/dx = 7
Problem 2 :
y = 2x³ - 4 (3, 50)
Solution :
y = 2x³ - 4
dy/dx = 2(3x²) - 0
dy/dx = 6x²
Put x = 3,
dy/dx = 6(3)²
dy/dx = 54
Problem 3 :
y = -x² + 1/x (-2, -4.5)
Solution :
y = -x² + 1/x
y = -x² + (x-1)
dy/dx = -2x - x(-1-1)
= -2x - x-2
dy/dx = -2x - 1/x2
Put x = -2,
dy/dx = -2(-2) - 1/(-2)²
= 4 - 1/4
dy/dx = 15/4
Find the equation of the tangent to the curve at the point indicated:
Problem 4 :
y = 3x² - x (1, 2)
Solution :
y = 3x² - x
dy/dx = 6x - 1
Put x = 1,
dy/dx = 6(1) - 1
dy/dx = 5
Equation of tangent:
y - y1 = m(x - x1)
y - 2 = 5(x - 1)
y - 2 = 5x - 5
y = 5x - 5 + 2
y = 5x - 3
Problem 5 :
y = x³ + 4x (-1, -5)
Solution :
y = x³ + 4x
dy/dx = 3x² + 4
Put x = 1,
dy/dx = 3(1)² + 4
dy/dx = 7
Equation of tangent:
y - y1 = m(x - x1)
y + 5 = 7(x + 1)
y + 5 = 7x + 7
y = 7x + 7 - 5
y = 7x + 2
Problem 6 :
y = x² - 1/x (1, 2)
Solution :
y = x² - 1/x
y= x² - (x-1)
dy/dx = 2x - (-x-2)
dy/dx = 2x + 1/x2
Put x = 1,
dy/dx = 2(1) + 1/(1)²
dy/dx = 3
Equation of tangent:
y - y1 = m(x - x1)
y - 2 = 3(x - 1)
y - 2 = 3x - 3
y = 3x - 3 + 2
y = 3x - 1
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM