Question

Differentiate (x²+1) (x²+3).

Answer 1

Answer:

4. If y = (sin-¹ x)², prove that (1-x²)y² - xy₁ - 2 = 0 and deduce that (1-x²) Yn+2 - (2n + 1)xYn+1 -n₂yn = 0.

Step-by-step explanation:

Answer 2

Required Solution :-

Firstly we'll multiply them.

• x² (x² + 3) + 1 (x² + 3)
• x⁴ + 3x² + x² + 3

Now, we'll differentiate them w.r.t. x.

We would be using the given theorum :

• d / dx (x^n) = nx^n-1

$\\ \implies \: \sf{ \dfrac{d}{dx}(x {}^{4}) \: + \: \dfrac{d}{dx}(3x {}^{2}) \: + \: \dfrac{d}{dx}(x {}^{2}) \: + \dfrac{d}{dx}(3)} \\ \\ \implies \: \sf{ \dfrac{d}{dx}(4x {}^{4 - 1}) \: + \: 3\dfrac{d}{dx}(x {}^{2}) \: + \: \dfrac{d}{dx}(x {}^{2}) \: + \dfrac{d}{dx}(3)} \\ \\ \implies \: \sf{ \dfrac{d}{dx}(4x {}^{4 - 1}) \: + \: 3\dfrac{d}{dx}(2x {}^{2 - 1}) \: + \: \dfrac{d}{dx}(2x {}^{2 - 1}) \: + \dfrac{d}{dx}(3)} \\ \\ \implies \: \sf{ \dfrac{d}{dx}(4x {}^{3}) \: + \: 3\dfrac{d}{dx}(2x {}^{2 - 1}) \: + \: \dfrac{d}{dx}(2x {}^{2 - 1}) \: + \dfrac{d}{dx}(3)} \\ \\ \implies \: \sf{ \dfrac{d}{dx}(4x {}^{3}) \: + \: 3\dfrac{d}{dx}(2x {}^{2 - 1}) \: + \: \dfrac{d}{dx}(2x {}^{}) \: + \dfrac{d}{dx}(3)} \\ \\ \implies \: \sf{ \dfrac{d}{dx}(4x {}^{3}) \: + \: 3\dfrac{d}{dx}(2x {}^{}) \: + \: \dfrac{d}{dx}(2x {}^{}) \: + \dfrac{d}{dx}(3)} \\ \\ \implies \: \sf{ \dfrac{d}{dx}(4x {}^{3}) \: + \: 3\dfrac{d}{dx}(2x {}^{}) \: + \: \dfrac{d}{dx}(2x {}^{}) \: + \dfrac{d}{dx}(0)} \\ \\ \implies \: \bf{ \red{4x {}^{3} \: + \: 6x \: + \: 2x}}$