APPLICATIONS OF DIFFERENTIATION A SUMMARY Applications of Differentiation

Applications of Differentiation • Find gradient of tangent; equation of tangent and normal •

Results (stationary points) For stationary point, dy/dx = 0. First Derivative Test (if stationary

Results (stationary points) For stationary point, dy/dx = 0. Second Derivative Test (if stationary

Results (max & min. values) For stationary point, d. A/dx = 0 (e. g)

Connected rates of change For stationary point, dy/dx = 0. it’s good to use

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APPLICATIONS OF DIFFERENTIATION (A SUMMARY)

Applications of Differentiation • Find gradient of tangent; equation of tangent and normal • Stationary points, max. & min. points, points of inflexion • Problems of maxima & minima

Results (stationary points) For stationary point, dy/dx = 0. First Derivative Test (if stationary point at x = a) (i) If dy/dx changes from negative to positive as x increases through a, the point is a minimum point. (ii) If dy/dx changes from positive to negative as x increases through a, the point is a maximum point. (iii) If dy/dx does not change sign as x increases through a, the point is a stationary point of inflexion.

Results (stationary points) For stationary point, dy/dx = 0. Second Derivative Test (if stationary point at x = a) (i) If (ii) If , the point is a minimum point. , the point is a maximum point. (iii) If & dy/dx does not change sign as x increases through a, the point is a stationary point of inflexion.

Results (max & min. values) For stationary point, d. A/dx = 0 (e. g) Second Derivative Test (if stationary value at x = a) (i) If (ii) If , the point is a minimum value. , the point is a maximum value.

Connected rates of change For stationary point, dy/dx = 0. it’s good to use chain rule to solve. . for example,