a) y=(x2+5x=4)cos3x
P.S x2 means x to the power of 2
b) find the euqation of the tangent to the curve at the point where x=0
can u solve a A level maths equation for me please URGENT
2008-09-04 10:44 am
回答 (1)
2008-09-04 3:44 pm
✔ 最佳答案
y = (x^2 + 5x - 4)cos3xdy/dx = cos3x(2x + 5) - 3(x^2 + 5x - 4)sin3x
When x = 0, dy/dx = 5 and y = -4. Therefore, equation of tangent at point (0,-4) is
y + 4 = 5( x - 0)
y + 4 = 5x
y = 5x - 4.
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