Commerce (English Medium) इयत्ता १२ - CBSE Question Bank Solutions for Mathematics

Differentiate the function with respect to x. 

cos x³ . sin² (x⁵)

Differentiate the function with respect to x. 

`2sqrt(cot(x^2))`

Differentiate the function with respect to x.

`cos (sqrtx)`

Prove that the function f given by f(x) = |x − 1|, x ∈ R is not differentiable at x = 1.

Differentiate the function with respect to x:

(3x² – 9x + 5)⁹

Differentiate the function with respect to x:

sin³ x + cos⁶ x

Differentiate the function with respect to x:

`(5x)^(3cos 2x)`

Differentiate the function with respect to x:

`sin^(–1)(xsqrtx), 0 ≤ x ≤ 1`

Differentiate the function with respect to x:

`(cos^(-1)  x/2)/sqrt(2x+7)`, −2 < x < 2

Differentiate the function with respect to x:

`x^(x^2 -3) + (x -3)^(x^2)`, for x > 3

Find `dy/dx`, if y = 12 (1 – cos t), x = 10 (t – sin t), `-pi/2 < t < pi/2`.

If (x – a)² + (y – b)² = c², for some c > 0, prove that `[1+ (dy/dx)^2]^(3/2)/((d^2y)/dx^2)` is a constant independent of a and b.

If f(x) = |x|³, show that f"(x) exists for all real x and find it.

Does there exist a function which is continuos everywhere but not differentiable at exactly two points? Justify your answer?

If y = `[(f(x), g(x), h(x)),(l, m,n),(a,b,c)]`, prove that `dy/dx = |(f'(x), g'(x), h'(x)),(l,m, n),(a,b,c)|`.

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = e^x + 1  :  y″ – y′ = 0

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = x² + 2x + C  :  y′ – 2x – 2 = 0

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = cos x + C : y′ + sin x = 0

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

`y sqrt(1 + x^2) : y' = (xy)/(1+x^2)`

Verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

y = Ax : xy′ = y (x ≠ 0)