If x = 2 a t 2 , y = a t 4 , then find d y d x .

Given that, x = 2 a t 2 , y = a t 4 . This is called implicit differentiation or indirect differentiation. Here, both the given terms are differentiated independently with respect to t , third variable and then combined. Differentiating x and y with respect to t we get, d x d t = 4 a t , d y d t = 4 a t 3 . Dividing d y d t by d x d t we get, d y d x = d y d t × d t d x = 4 a t 3 4 a t = t 2 . Thus the value of d y d x is t 2 .

If x = 2 a t 2 , y = a t 4 , then find d y d x .

Given that, x = 2 a t 2 , y = a t 4 . This is called implicit differentiation or indirect differentiation. Here, both the given terms are differentiated independently with respect to t , third variable and then combined. Differentiating x and y with respect to t we get, d x d t = 4 a t , d y d t = 4 a t 3 . Dividing d y d t by d x d t we get, d y d x = d y d t × d t d x = 4 a t 3 4 a t = t 2 . Thus the value of d y d x is t 2 .

Differentiate sec ( tan x ) with respect to x .

Given that, the function is sec ( tan x ) . Differentiating with respect to x , we get d d x sec ( tan x ) = sec ( tan x ) tan ( tan x ) d d x tan x ⇒ d d x sec ( tan x ) = sec ( tan x ) tan ( tan x ) sec 2 x d d x x ⇒ d d x sec ( tan x ) = sec ( tan x ) tan ( tan x ) sec 2 x 1 2 x Thus the required value is sec ( tan x ) tan ( tan x ) sec 2 x 2 x .

If x = a ( cos θ + θ sin θ ) and y = a ( sin θ - θ cos θ ) , then find d y d x , at θ = π 4 .

We have x = a ( cos θ + θ sin θ ) Differentiate w.r.t θ , we get d x d θ = a d d θ cos θ + a d d θ θ sin θ ⇒ d x d θ = - a sin θ + a θ d d θ sin θ + sin θ d d θ θ [Using product rule of differentiation] ⇒ d x d θ = - a sin θ + a θ cos θ + sin θ ⇒ d x d θ = a θ cos θ and y = a ( sin θ - θ cos θ ) Differentiate w.r.t θ , we get: d y d θ = a d d θ sin θ - a d d θ θ cos θ ⇒ d y d θ = a cos θ - a θ d d θ cos θ + cos θ d d θ θ ⇒ d y d θ = a cos θ - a - θ sin θ + cos θ ⇒ d y d θ = a θ sin θ Now, d y d θ d x d θ = a θ sin θ a θ cos θ ⇒ d y d x = tan θ Thus, d y d x at θ = π 4 is tan π 4 = 1 .

If y = ( sin x ) sin x , 0 < x < π , find d y d x .

Given that, y = ( sin x ) sin x , 0 < x < π Applying logarithm on both sides we get log y = sin x · log ( sin x ) Differentiating on both sides with respect to x 1 y d y d x = sin x · 1 sin x cos x + log sin x · cos x ⇒ 1 y d y d x = cos x + log sin x · cos x ⇒ 1 y d y d x = cos x 1 + log sin x ⇒ d y d x = ( sin x ) sin x cos x 1 + log sin x

If y = ( sin x ) sin x , 0 < x < π , find d y d x .

Given that, y = ( sin x ) sin x , 0 < x < π Applying logarithm on both sides we get log y = sin x · log ( sin x ) Differentiating on both sides with respect to x 1 y d y d x = sin x · 1 sin x cos x + log sin x · cos x ⇒ 1 y d y d x = cos x + log sin x · cos x ⇒ 1 y d y d x = cos x 1 + log sin x ⇒ d y d x = ( sin x ) sin x cos x 1 + log sin x

If x = a ( cos θ + θ sin θ ) and y = a ( sin θ - θ cos θ ) , then find d y d x , at θ = π 4 .

We have x = a ( cos θ + θ sin θ ) Differentiate w.r.t θ , we get d x d θ = a d d θ cos θ + a d d θ θ sin θ ⇒ d x d θ = - a sin θ + a θ d d θ sin θ + sin θ d d θ θ [Using product rule of differentiation] ⇒ d x d θ = - a sin θ + a θ cos θ + sin θ ⇒ d x d θ = a θ cos θ and y = a ( sin θ - θ cos θ ) Differentiate w.r.t θ , we get: d y d θ = a d d θ sin θ - a d d θ θ cos θ ⇒ d y d θ = a cos θ - a θ d d θ cos θ + cos θ d d θ θ ⇒ d y d θ = a cos θ - a - θ sin θ + cos θ ⇒ d y d θ = a θ sin θ Now, d y d θ d x d θ = a θ sin θ a θ cos θ ⇒ d y d x = tan θ Thus, d y d x at θ = π 4 is tan π 4 = 1 .

Differentiate sec ( tan x ) with respect to x .

Given that, the function is sec ( tan x ) . Differentiating with respect to x , we get d d x sec ( tan x ) = sec ( tan x ) tan ( tan x ) d d x tan x ⇒ d d x sec ( tan x ) = sec ( tan x ) tan ( tan x ) sec 2 x d d x x ⇒ d d x sec ( tan x ) = sec ( tan x ) tan ( tan x ) sec 2 x 1 2 x Thus the required value is sec ( tan x ) tan ( tan x ) sec 2 x 2 x .

Find d y d x from the equation 2 x + 3 y = sin y .

Given that, the equation is 2 x + 3 y = sin y Differentiating with respect to x ⇒ 2 + 3 d y d x = cos y d y d x ⇒ 2 = cos y - 3 d y d x ⇒ d y d x = 2 cos y - 3 Thus the value of d y d x is 2 cos y - 3 .

Differentiate sin x 2 + 5 w.r.t x .

To differentiate sin x 2 + 5 w.r.t x : So, d d x sin x 2 + 5 = cos x 2 + 5 × d d x x 2 + 5 ∵ d d x sin f x = cos f x × f ' x ⇒ d d x sin x 2 + 5 = cos x 2 + 5 × d d x x 2 + d d x 5 ∵ d d x x n = n x n - 1 ⇒ d d x sin x 2 + 5 = cos x 2 + 5 × 2 x Therefore, d d x sin x 2 + 5 = 2 x cos x 2 + 5 .

Show that the function f ( x ) = | x - 2 | , x ∈ R is continuous but not differentiable at x = 2 .

Given that, the function is f ( x ) = | x - 2 | , x ∈ R We can redefine the function as f x = - ( x - 2 ) ; if x < 2 x - 2 ; if x ≥ 2 Now, continuity at x = 2 So, LHL = lim x → 2 + f x = lim h → 0 f 2 + h = lim h → 0 | 2 + h - 2 | = lim h → 0 | h | = 0 Similarly RHL = lim x → 2 - f x = lim h → 0 f 4 - h = lim h → 0 | 4 - h - 4 | = lim h → 0 | - h | = 0 and also, f 2 = 0 Thus, clearly, LHL = RHL = f 2 Hence, f x is continuous at x = 2 For Differentiability at x = 2 , Now, LHD = lim x → 2 - f ( x ) - f ( 2 ) x - 2 = lim x → 2 - - ( x - 2 ) - 0 x - 2 = - 1 And similarly, RHD = lim x → 2 + f ( x ) - f ( 2 ) x - 2 = lim x → 2 + x - 2 - 0 x - 2 = 1 Thus, LHD ≠ RHD Hence, f x is not differentiable at x = 2 .