NTA Abhyas JEE Main2020MathematicsDifferentiationPractice

If y = t a n - 1 x 1 + 6 x 2 + t a n - 1 2 x - 1 2 x + 1 ∀ x > 0 , then d y d x is equal to

Options

A3 1 + 9 x 2
B1 1 + 6 x 2
C1 1 + 6 x 2 + 1 1 + x 2
D3 1 + 6 x 2

Correct answer

A. 3 1 + 9 x 2

Step-by-step solution

y = t a n - 1 3 x - 2 x 1 + 3 x ⋅ 2 x + t a n - 1 2 x - 1 1 + 2 x ⋅ 1 y = t a n - 1 3 x - t a n - 1 2 x + t a n - 1 2 x - tan - 1 1 y = t a n - 1 3 x - π 4 ⇒ d y d x = 3 1 + 9 x 2

Practice Differentiation on Quantrex Academy →

More from Differentiation

Let R denote the set of all real numbers. Consider the polynomial function f: R R defined by f(x) = d¹⁰ dx¹⁰ ((x^2 - 1)¹⁰ ) , for all x R . Here d¹⁰ dx¹⁰ ((x^2 - 1)¹⁰ ) is the 10th 2026 Let f(x) and g(x) be twice differentiable functions satisfying f''(x) = g''(x) for all x R , f'(1) = 2g'(1) = 4 and g(2) = 3f(2) = 9 . Then f(25) - g(25) is equal to : 2026 Let f be a real polynomial of degree n such that f(x) = f'(x) f''(x) , for all x R . If f(0) = 0 , then 36 (f'(2) + f''(2) + ₀^2 f(x) ,dx ) is equal to: 2026 Let f(x)=x³+x² f^ (1)+2 x f^ (2)+f^ (3), x R . Then the value of f^ (5) is: 2026 Let R denote the set of all real numbers. Let f: R R and g: R (0,4) be functions defined by f(x)= _e (x^2+2 x+4 ) , and g(x)= 4 1+e^ -2 x Define the composite function f g⁻¹ by (f 2025 Let f: R R be a twice differentiable function such that ( x y)(f(2 x+2 y)-f(2 x-2 y))=( x y )(f(2 x +2 y )+f(2 x -2 y )) , for all x , y R . If f^ (0)= 1 2 , then the value of 24 f 2025 If _e y=3 ⁻¹ x , then (1-x^2 ) y^ -x y^ at x= 1 2 is equal to 2024 Let f(x)=a x^3+b x^2+c x+41 be such that f(1)=40, f^ (1)=2 and f^ (1)=4 . Then a ^2+ b ^2+ c ^2 is equal to: 2024 Full Differentiation list All NTA Abhyas JEE Main PYQs