Chapter 5
sin (x2 + 5) 2. cos (sin x) 3. sin (ax + b)
sec (tan ( x )) 5. / sin ( ) / cos ( ) ax b cx d + + 6. cos x3 . sin 2 (x5) 7. ( )22 cot x 8. ( )cos x 9. Prove that the function f given by f (x) = |x - 1|, x ∈ R is not differentiable at x = 1. 10. Prove that the greatest integer function defined by f (x) = [x], 0 < x < 3 is not differentiable at x = 1 and x = 2. 5.3.2 Derivatives of implicit functions Until now we have been differentiating various functions given in the form y = f (x). But it is not necessary that functions are always expressed in this form. For example, consider one of the following relationships between x and y: x - y - π = 0 x + sin xy - y = 0 In the first case, we can solve for y and rewrite the relationship as y = x - π. In the second case, it does not seem that there is an easy way to solve for y. Nevertheless, there is no doubt about the dependence of y on x in either of the cases. When a relationship between x and y is expressed in a way that it is easy to solve for y and write y = f (x), we say that y is given as an explicit function of x. In the latter case it is implicit that y is a function of x and we say that the relationship of the second type, above, gives function implicitly. In this subsection, we learn to dif ferentiate implicit functions. Example 22 Find dy dx if x - y = π. Solution One way is to solve for y and rewrite the above as y = x - π But then dy dx = 1 Alternatively, directly differentiating the relationship w.r.t., x, we have
( )d x ydx − = d dx π Recall that d dx π means to differentiate the constant function taking value π everywhere w.r.t., x. Thus ( ) ( )d d x ydx dx − = 0 which implies that dy dx = 1dx dx = Example 23 Find dy dx , if y + sin y = cos x. Solution We differentiate the relationship directly with respect to x, i.e., (sin )dy d ydx dx+ = (cos )d xdx which implies using chain rule cosdy dy ydx dx+ ⋅ = - sin x This gives dy dx = sin 1 cos x y− + where y ≠ (2n + 1) π MATHEMATICS124 5.3.3 Derivatives of inverse trigonometric functions We remark that inverse trigonometric functions are continuous functions, but we will not prove this. Now we use chain rule to find derivatives of these functions. Example 24 Find the derivative of f given by f (x) = sin-1 x assuming it exists. Solution Let y = sin-1 x. Then, x = sin y. Differentiating both sides w.r.t. x, we get 1 = cos y dy dx which implies that dy dx = 1 1 1 cos cos(sin )y x−= Observe that this is defined only for cos y ≠ 0, i.e., sin-1 x ≠ ,2 2 π π− , i.e., x ≠ - 1, 1, i.e., x ∈ (- 1, 1). To make this result a bit more attractive, we carry out the following manipulation. Recall that for x ∈ (- 1, 1), sin (sin -1 x) = x and hence cos² y = 1 - (sin y)2 = 1 - (sin (sin -1 x))2 = 1 - x2 Also, since y ∈ ,2 2 π π − , cos y is positive and hence cos y = 21 x− Thus, for x ∈ (- 1, 1), 2 1 1 cos 1 dy
| dx | y | x |
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= = / − 2/1 / 1 x− 2/1 / 1 x / − / − 21 / 1 x+ f (x) sin -1 x cos -1 x tan -1x Domain off (-1, 1) (-1, 1) R f 1(x)