✔ 最佳答案
Q1.
Step 1. Find dy/dx of the function.
y = 3x^2 - 4
dy/dx = 6x = slope of tangent.
For point (1, -1), slope = 6(1) = 6.
Step 2. Use equation of straight line to find the tangent,
y - (-1) = 6( x - 1)
y + 1 = 6x - 6
y = 6x - 7 is the tangent at (1,-1) of the curve.
Same method for Q3 and Q4.
Q2.
dq/dx = 1/3sqrt 8 ( 2x) = 2x/3(2sqrt 2) = x/3sqrt2.
2009-09-23 12:07:51 補充：
Q2. if q = (8x^2)^(-1/3), by Chain Rule, dq/dx = [d(8x^2)^(-1/3)/d(8x^2)][d(8x^2)/dx] = [(-1/3)(8x^2)^(-4/3)][16x] = (-1/3)( 8)^(-4/3)(x^2)^(-4/3)(16x)
= (-1/3)(2)^(- 4)(x)^(-8/3)(16x) = (-1/3)(1/16)(x)^(-8/3)(16x) = (-1/3)(x)^(-5/3).
2009-09-23 12:11:43 補充：
Q3. dy/dx = 8x + 5. Slope at point (1,15) = 8(1) + 5 = 13. eqt of tangent is y - 15 = 13(x - 1). y - 15 = 13x - 13, that is y = 13x + 2.
2009-09-23 12:14:03 補充：
Q4. dy/dx = -2x/5. Slope at point ( 4,-3) = -2(4)/5 = -8/5. Eqt. of tangent is y + 3 = -8/5(x - 4). 5y + 15 = -8x + 32, that is 5y + 8x - 17 = 0.
