Notes On Derivatives of composite, implicit and inverse trigonometric functions - CBSE Class 12 Maths
Derivatives of Composite, Implicit and Inverse Trigonometric Functions. Chain rule: f is a real valued function, which is a composite of two functions, u and v; i.e. f = u • v. Suppose t = v(x) and if dt/dx and du/dt exist, then: df/dx = du/dt • dt/dx The chain rule can be also be applied to functions that are composites of more than two functions. f is a real valued function, which is a composite of three functions, u, v and w; i.e. f = (u • v) • w. If s = w(x) and t = v(s), then: df/dx = d(u • v)/ds • ds/dx = du//dt • dt/ds • ds/dx Exmple : f(s) = (x - 3)50 Sol: Let f(x) = (h • g)(x) Where g(x) = x - 3 and h(x) = x50 Suppose t = x - 3 ⇒ f(x) = h(t) = t50 df/dx = dh/dt • dt/dx = d(t50)/dt • d(x - 3)/dx = 50t49 × 1 = 50t49 Substitute the value of t. ∴ df/dx = 50(x - 3)49 An expression involving x and y that is easily solved for y and written in the form y = f(x) is said to be y is given as an explicit function of x. Ex: y = f(x) = 2x + 5 A function that is formed implicitly is called an implicit function. y - 2x = 5 Inverse Trigonometric Functions      f(x)     sin-1 x     cos-1 x    tan-1 x    cot-1 x     sec-1 x    cosex-1 x     f ' (x)    1/√(1-x2)    -1/√(1-x2)     1/1+x2   -1/1+x2   1/(x√(x2 - 1))    -1/(x√(x2 - 1)) Domain of f '     (-1, 1)     (-1, 1)      R     R  (-∞, -1) ∪ (1, ∞)  (-∞, -1) ∪ (1, ∞) Ex: f(x) = sin-1x Let y = sin-1x ⇒ x = sin y ..... (1) dx/dx = d(sin y)/dx ⇒ 1 = cos y dy/dx ⇒ dy/dx = 1/cos y = 1/√(1 - sin2 y) = 1/√(1 - sin y)2 [ '.' cos2y + sin2 y = 1] = 1/ √(1 - x2) [From (1)] ∴ dy/dx = 1/√(1 - x2) Note: The derivative of sin-1 x is only possible for x ∈ (-1, 1) Similarly we can find all the derivatives of inverse trigonometric functions.

#### Summary

Derivatives of Composite, Implicit and Inverse Trigonometric Functions. Chain rule: f is a real valued function, which is a composite of two functions, u and v; i.e. f = u • v. Suppose t = v(x) and if dt/dx and du/dt exist, then: df/dx = du/dt • dt/dx The chain rule can be also be applied to functions that are composites of more than two functions. f is a real valued function, which is a composite of three functions, u, v and w; i.e. f = (u • v) • w. If s = w(x) and t = v(s), then: df/dx = d(u • v)/ds • ds/dx = du//dt • dt/ds • ds/dx Exmple : f(s) = (x - 3)50 Sol: Let f(x) = (h • g)(x) Where g(x) = x - 3 and h(x) = x50 Suppose t = x - 3 ⇒ f(x) = h(t) = t50 df/dx = dh/dt • dt/dx = d(t50)/dt • d(x - 3)/dx = 50t49 × 1 = 50t49 Substitute the value of t. ∴ df/dx = 50(x - 3)49 An expression involving x and y that is easily solved for y and written in the form y = f(x) is said to be y is given as an explicit function of x. Ex: y = f(x) = 2x + 5 A function that is formed implicitly is called an implicit function. y - 2x = 5 Inverse Trigonometric Functions      f(x)     sin-1 x     cos-1 x    tan-1 x    cot-1 x     sec-1 x    cosex-1 x     f ' (x)    1/√(1-x2)    -1/√(1-x2)     1/1+x2   -1/1+x2   1/(x√(x2 - 1))    -1/(x√(x2 - 1)) Domain of f '     (-1, 1)     (-1, 1)      R     R  (-∞, -1) ∪ (1, ∞)  (-∞, -1) ∪ (1, ∞) Ex: f(x) = sin-1x Let y = sin-1x ⇒ x = sin y ..... (1) dx/dx = d(sin y)/dx ⇒ 1 = cos y dy/dx ⇒ dy/dx = 1/cos y = 1/√(1 - sin2 y) = 1/√(1 - sin y)2 [ '.' cos2y + sin2 y = 1] = 1/ √(1 - x2) [From (1)] ∴ dy/dx = 1/√(1 - x2) Note: The derivative of sin-1 x is only possible for x ∈ (-1, 1) Similarly we can find all the derivatives of inverse trigonometric functions.

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