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NOTE: 1) If f(x) = x^n ; where n = integer; then
f'(x) = d[f(x)]/dx = n*x^(n-1)
2) If f(x) = k ; where 'k' = real number, then
f'(x) = d[f(x)]/dx = 0
Equation of curve :-
y = x^4 + 2*(x^2) - x
Differentiating the given fuction with respect to x we get
y'(x) = 4*(x^3) + 4*(x^1) - 1*(x^0) = 4*(x^3) + 4x -1
Now, y'(x) at (1,2) = slope of the tangent to the curve at (1,2)
or, y(1) = 4 + 4 - 1 = 7
Now, equation of line passing through a point (a,b) and having a slope 's' is given by the equation :-
y-b = s*(x-a)
Thus, equation of the tangent to the curve at (1,2) is :-
y-2 = 7*(x-1)
or, y-2 = 7x - 7
or, y - 7x + 5 = 0 is the equation of the tangent to the given curve.
Given curve is y=x^4+2x^2-x
The slope of the tangent at a point is equal to the derivative of the curve at that point.
The slope of the tangent at (1,2) is 7
The equation of the tangent at (1,2) is y-2=7(x-1)
That is y=7x - 5.
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