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Math  /  Algebra

Question4. y=(x2+1)y = (x^2 + 1) arccot xx 9

Studdy Solution

STEP 1

What is this asking? We need to find the derivative of a function that's a product of a polynomial and an inverse cotangent function! Watch out! Don't forget the chain rule and the product rule, and remember the derivative of arccot(x)!

STEP 2

1. Define the function
2. Recall important derivatives
3. Apply the product rule
4. Simplify the derivative

STEP 3

Let's **define** our function clearly: y=(x2+1)⋅arccot(x).y = (x^2 + 1) \cdot \text{arccot}(x).

STEP 4

We'll need the **power rule** for the derivative of x2+1x^2 + 1: ddx(x2+1)=2x.\frac{d}{dx}(x^2 + 1) = 2x.

STEP 5

And the derivative of **arccotangent**: ddx(arccot(x))=−11+x2.\frac{d}{dx}(\text{arccot}(x)) = -\frac{1}{1 + x^2}.

STEP 6

Remember the **product rule**: (uv)′=u′v+uv′(uv)' = u'v + uv'.
Here, u=x2+1u = x^2 + 1 and v=arccot(x)v = \text{arccot}(x).

STEP 7

Let's apply it! dydx=ddx((x2+1)⋅arccot(x))\frac{dy}{dx} = \frac{d}{dx}((x^2 + 1) \cdot \text{arccot}(x)) dydx=(2x)⋅arccot(x)+(x2+1)⋅(−11+x2).\frac{dy}{dx} = (2x) \cdot \text{arccot}(x) + (x^2 + 1) \cdot \left(-\frac{1}{1 + x^2}\right).

STEP 8

Now, let's **simplify**!
Notice that (x2+1)(x^2 + 1) appears in both the numerator and denominator of the second term.
We can divide both the numerator and denominator by (x2+1)(x^2 + 1) to simplify it. dydx=2x⋅arccot(x)−x2+1x2+1.\frac{dy}{dx} = 2x \cdot \text{arccot}(x) - \frac{x^2 + 1}{x^2 + 1}. dydx=2x⋅arccot(x)−1.\frac{dy}{dx} = 2x \cdot \text{arccot}(x) - 1.

STEP 9

The **derivative** of y=(x2+1)arccot(x)y = (x^2 + 1)\text{arccot}(x) is 2x⋅arccot(x)−12x \cdot \text{arccot}(x) - 1.

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