06Class 12 Maths

Application of Derivatives formulas

Rates of change, tangents, maxima and minima. Word problems here always come down to writing one function and differentiating it.

Rates and tangents

Rate of change
dydt=dydx⋅dxdt\displaystyle \frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt}
Slope of the tangent
m=dydx∣(x0,y0)\displaystyle m = \left.\frac{dy}{dx}\right|_{(x_0, y_0)}
Tangent line
y−y0=m(x−x0)\displaystyle y - y_0 = m(x - x_0)
Normal line
y−y0=−1m(x−x0)\displaystyle y - y_0 = -\frac{1}{m}(x - x_0)

Increasing and decreasing

Increasing
f′(x)>0\displaystyle f'(x) > 0
Decreasing
f′(x)<0\displaystyle f'(x) < 0

Maxima and minima

Critical point
f′(c)=0 or f′(c) does not exist\displaystyle f'(c) = 0 \ \text{or} \ f'(c) \ \text{does not exist}
Local maximum
f′(c)=0, f′′(c)<0\displaystyle f'(c) = 0, \ f''(c) < 0
Local minimum
f′(c)=0, f′′(c)>0\displaystyle f'(c) = 0, \ f''(c) > 0

If f''(c) = 0, use the first derivative test instead.

Where students slip up

  • Forgetting endpoints when finding an absolute maximum on a closed interval.
  • Assuming f'(c) = 0 always means a maximum or minimum. x³ at 0 is neither.
  • Using the tangent's slope for the normal without taking the negative reciprocal.

Try it on your weakest chapter.

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