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 â€¢ v. Suppose t = v(x) and if dt/dx and du/dt exist, then: df/dx = du/dt â€¢ 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 â€¢ v) â€¢ w. If s = w(x) and t = v(s), then:
df/dx = d(u â€¢ v)/ds â€¢ ds/dx = du//dt â€¢ dt/ds â€¢ ds/dx
Exmple :
f(s) = (x - 3)^{50}
Sol:
Let f(x) = (h â€¢ g)(x) Where g(x) = x - 3 and h(x) = x^{50}
Suppose t = x - 3 â‡’ f(x) = h(t) = t^{50}
df/dx = dh/dt â€¢ dt/dx
= d(t^{50})/dt â€¢ d(x - 3)/dx
= 50t^{49} Ã— 1 = 50t^{49} Substitute the value of t.
âˆ´ 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/âˆš(1-x^{2}) |
-1/âˆš(1-x^{2}) |
1/1+x^{2} |
-1/1+x^{2} |
1/(xâˆš(x^{2} - 1)) |
-1/(xâˆš(x^{2} - 1)) |
Domain of f ' |
(-1, 1) |
(-1, 1) |
R |
R |
(-âˆž, -1) âˆª (1, âˆž) |
(-âˆž, -1) âˆª (1, âˆž) |
Ex: f(x) = sin^{-1}x
Let y = sin^{-1}x
â‡’ x = sin y ..... (1)
dx/dx = d(sin y)/dx â‡’ 1 = cos y dy/dx
â‡’ dy/dx = 1/cos y
= 1/âˆš(1 - sin^{2} y) = 1/âˆš(1 - sin y)^{2} [ '.' cos^{2}y + sin^{2} y = 1]
= 1/ âˆš(1 - x^{2}) [From (1)]
âˆ´ dy/dx = 1/âˆš(1 - x^{2})
Note:
The derivative of sin^{-1} x is only possible for x âˆˆ (-1, 1)
Similarly we can find all the derivatives of inverse trigonometric functions.
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 â€¢ v. Suppose t = v(x) and if dt/dx and du/dt exist, then: df/dx = du/dt â€¢ 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 â€¢ v) â€¢ w. If s = w(x) and t = v(s), then:
df/dx = d(u â€¢ v)/ds â€¢ ds/dx = du//dt â€¢ dt/ds â€¢ ds/dx
Exmple :
f(s) = (x - 3)^{50}
Sol:
Let f(x) = (h â€¢ g)(x) Where g(x) = x - 3 and h(x) = x^{50}
Suppose t = x - 3 â‡’ f(x) = h(t) = t^{50}
df/dx = dh/dt â€¢ dt/dx
= d(t^{50})/dt â€¢ d(x - 3)/dx
= 50t^{49} Ã— 1 = 50t^{49} Substitute the value of t.
âˆ´ 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/âˆš(1-x^{2}) |
-1/âˆš(1-x^{2}) |
1/1+x^{2} |
-1/1+x^{2} |
1/(xâˆš(x^{2} - 1)) |
-1/(xâˆš(x^{2} - 1)) |
Domain of f ' |
(-1, 1) |
(-1, 1) |
R |
R |
(-âˆž, -1) âˆª (1, âˆž) |
(-âˆž, -1) âˆª (1, âˆž) |
Ex: f(x) = sin^{-1}x
Let y = sin^{-1}x
â‡’ x = sin y ..... (1)
dx/dx = d(sin y)/dx â‡’ 1 = cos y dy/dx
â‡’ dy/dx = 1/cos y
= 1/âˆš(1 - sin^{2} y) = 1/âˆš(1 - sin y)^{2} [ '.' cos^{2}y + sin^{2} y = 1]
= 1/ âˆš(1 - x^{2}) [From (1)]
âˆ´ dy/dx = 1/âˆš(1 - x^{2})
Note:
The derivative of sin^{-1} x is only possible for x âˆˆ (-1, 1)
Similarly we can find all the derivatives of inverse trigonometric functions.