Quiz 3
Quiz 3
Friday, March 30, 2001
NAME:
Student Number:_{}
No._{} Marks _{}
(20 Marks) Find the equation of the tangent line to
y = [(3x5)/(x^{2}+7)] at the point (0, 5/7).
Solution.



(3x5)¢(x^{2}+7)(3x5)(x^{2}+7)¢ (x^{2}+7)^{2}


 


(3(x^{2}+7)(3x5)(2x) (x^{2}+7)^{2}


 


3x^{2}+10x +21 (x^{2}+7)^{2}




Hence we have
y¢(0) = [21/47] = [3/7].
The equation of the tangent line is
y = (5/7)+y¢(0)(x0) = 5/7+(3/7)x 

or
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On 2 Apr 2001, 14:18.