Differential Calculus and its Applications
John Bird · 2008
Thus dy dx ( )(4)x ( )( )x ( )x ( ) x 5 4 1 1 2 2 1 1 2 ⎞ ⎠ ⎟⎟⎟ 20 4 1 2 3 3 32x x x i.e. dy dx 20x 4 x 2 x 1 1 dy dx 1 x Application: Differentiate y x x x x 5 4 1 2 1 with respect to x When y 3 sin 4x 2 cos 3x then dy dx (3)(4 cos 4x) (2)( 3 sin 3x) 12 cos 4x 6 sin 3x Application: Find the differential coefficient of y 3 sin 4x 2 cos 3x Application: Determine the derivative of f( ) e θ θ θ 6 2 3 ln f( ) e 2 eθ θ θ 2 6 2 6 2 3 3ln ln Hence, f ( ) ( )( )e e θ θ θ θ θ2 3 6 1 6 63 3⎛ ⎝ ⎜⎜⎜ ⎞ ⎠ ⎟⎟⎟ 6 6 e3θ θ When y uv, and u and v are both functions of x, then: dy dx u dv dx v du dx When y u v , and u and v are both functions of x then: dy dx v du dx u dv dx v2 Application: Find the differential coefficient of y 3x 2 sin 2x 3x 2 sin 2x is a product of two terms 3x 2 and sin 2x Let u 3x 2 and v sin 2x Using the product rule: dy dx u dv dx v du dx ↓ ↓ ↓ ↓ gives: dy dx ( x )( x) ( x)( ) 3 2 2 2 62 cos sin x i.e. dy dx x x 6 6 22 cos sin2x x 6x x 2x sin 2x( cos ) Application: Find the differential coefficient of y x x 4 5 5 4 sin 4 5 5 4 sin x x is a quotient. Let u 4 sin 5x and v 5x 4 dy dx v du dx u dv dx v x x x x x 5 20 5 4 5 20 5 ( )( cos ) ( sin )( ) ( ) x x x x x x x x x x 5 80 5 25 20 5 5 4 5 25 cos sin [ cos sin ] i.e. dy dx 4 5x (5x cos 5x 4 sin 5x) 5 Application: Determine the differential coefficient of y tan ax y tan ax sin cos ax ax . Differentiation of tan ax is thus treated as a quotient with u sin ax and v cos ax dy dx v du dx u dv dx v ax a ax ax a ax ax (cos )( cos ) (sin )( sin ) (cos )2 a ax a ax ax a ax ax ax a ax cos sin (cos ) (cos sin ) cos cos since cos sin2 2 1ax ax Hence, dy dx a sec 2 ax since sec 2 ax 1 2cos ax It is often easier to make a substitution before differentiating.