Equation of a straight line which is tangent to a curve

For the curve f(x) = x - 1/3x^2 (one third x squared), find the equation of the straight line which is tangent to this curve at the point x = 1.

Equation of the curve= f(x) = x - 1/3x^2
y= x- 1/3 x ^2
Finding derivative is easy
Derivative of x ^n is n x ^ (n-1)
Thus derivative of x ^2 is
2 x ^(2-1)= 2 x ^ 1 that is 2x
Derivative of x is
x= x ^1
therefore derivative of x is 1 * x ^ 1-1
= 1 * X ^ 0
=1 ( as X^0=1)
Similarly derivative of 1/x^2 can be solved
1/x^2 can be written as x ^ -2
Derivative of x ^-2 is -2 * x ^ (-2-1)= -2 x ^ -3 = -2/ x ^3

If the value is multiplied/ divided by a constant multiply/ divide the derivative by the constant.
Thus derivative of x ^2 is 2x.
Derivative of 3 x ^2 is 3 *( 2 x)= 6x

Therefore derivative of the curve f(x)= y= x- 1/3 x ^2
dy/dx = 1- 1/3 (-2/ x ^3)=1+ 2/3 (1/x^3)

At x=1
dy/dx= 1+ 2/3(1/1^3)=1+2/3= 5/3

The y coordinate of the point on the curve where x=1 is
y= x- 1/3 x ^2
=1-1/3(1/1^2)=1-1/3=2/3 ( as you have rightly calculated)
Thus the coordinates through which the straight line passes is (1, 2/3)
Since the st line is tangent to this curve at x=1 the slope of the st line = slope of the curve we have calculated. = 5/3
Equation of a st line is y= mx+c
m= 5/3
Substituting the values of m, y, x in the equation we can calculate the value of c
y= mx+c
2/3=5/3 (1)+ c ( since the st line passes through (1, 2/3)
or 2/3=5/3 + c or c= 2/3-5/3=-3/3=-1
Therefore the equation of the st line is
y= 5/3 (x) - 1

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