Show that the function f x = x sin 1 x when x ≠ 0 0 when x = 0 , is continuous but not differentiable at x = 0 .

Given: f x = x sin 1 x when x ≠ 0 0 when x = 0 If g x is continuous at x = a then lim x → a g x = g a . Now, lim x → 0 f x = lim x → 0 x sin 1 x = 0 = f 0 Hence, f x is continuous at x = 0 . Now, f ' 0 = lim h → 0 f 0 + h - f 0 h ⇒ f ' 0 = lim h → 0 f h - 0 h ⇒ f ' 0 = lim h → 0 h sin 1 h h ⇒ f ' 0 = lim h → 0 sin 1 h f ' 0 is not defined since value of lim h → 0 sin 1 h lies between - 1 to 1 . Hence, f x is not differentiable at x = 0 .

If y = sin cos x 2 , find d y d x .

Given: y = sin cos x 2 Differentiating both sides w.r.t. x , we get d y d x = cos cos x 2 × d d x cos x 2 ⇒ d y d x = cos cos x 2 - sin x 2 d d x x 2 ⇒ d y d x = - 2 x sin x 2 cos cos x 2

If y = sin x + x 2 cot 2 x , find d y d x .

Given: y = sin x + x 2 cot 2 x Differentiating both sides w.r.t. x , we get d y d x = d d x sin x + x 2 × cot 2 x - d d x cot 2 x × sin x + x 2 cot 2 2 x ⇒ d y d x = cos x + 2 x cot 2 x - - 2 cosec 2 2 x sin x + x 2 cot 2 2 x ⇒ d y d x = cos x + 2 x cot 2 x + 2 cosec 2 2 x sin x + x 2 cot 2 2 x ⇒ d y d x = cos x + 2 x tan 2 x + 2 sin x + x 2 sec 2 2 x

If y = cos x + sin x cos x - sin x , show that d y d x = sec 2 x + π 4 .

Given: y = cos x + sin x cos x - sin x Differentiating both sides w.r.t. x , we get d y d x = ( cos x - sin x ) × ( - sin x + cos x ) - ( cos x + sin x ) × ( - sin x - cos x ) ( cos x - sin x ) 2 ⇒ d y d x = ( cos x - sin x ) 2 + ( cos x + sin x ) 2 ( cos x - sin x ) 2 ⇒ d y d x = cos 2 x + sin 2 x - 2 cos x sin x + cos 2 x + sin 2 x + 2 cos x sin x ( cos x - sin x ) 2 ⇒ d y d x = 2 cos 2 x + sin 2 x ( cos x - sin x ) 2 ⇒ d y d x = 2 ( cos x - sin x ) 2 ⇒ d y d x = 1 1 2 cos x - 1 2 cos x 2 ⇒ d y d x = 1 cos 2 x + π 4 ⇒ d y d x = sec 2 x + π 4

If x y = e x - y , prove that d y d x = log x 1 + log x 2 .

Given: x y = e x - y . Applying logarithm on both sides. y log x = x - y ⇒ y + y log x = x y = x 1 + log x On differentiating both sides with respect to x , we get d y d x = 1 + log x · 1 - x 1 x 1 + log x 2 ⇒ d y d x = 1 + log x - 1 1 + log x 2 ⇒ d y d x = log x 1 + log x 2 Hence, proved.