(0 Find the derivative dy dx tan-1 x2 y = x2 + 1
The Correct Answer and Explanation is:
To find the derivative dydx\frac{dy}{dx}dxdy of the given function y=tan−1(x2+1)y = \tan^{-1}(x^2 + 1)y=tan−1(x2+1), we’ll need to use implicit differentiation and the chain rule.
Step 1: Differentiation of the Inverse Tangent Function
The derivative of the inverse tangent function tan−1(u)\tan^{-1}(u)tan−1(u), where uuu is a function of xxx, is: ddx(tan−1(u))=11+u2⋅dudx\frac{d}{dx}\left(\tan^{-1}(u)\right) = \frac{1}{1 + u^2} \cdot \frac{du}{dx}dxd(tan−1(u))=1+u21⋅dxdu
In this case, u=x2+1u = x^2 + 1u=x2+1, so we need to differentiate u=x2+1u = x^2 + 1u=x2+1 with respect to xxx.
Step 2: Differentiate the Inner Function
The derivative of u=x2+1u = x^2 + 1u=x2+1 with respect to xxx is: dudx=2x\frac{du}{dx} = 2xdxdu=2x
Step 3: Apply the Chain Rule
Now we can apply the chain rule to differentiate the entire expression. We get: dydx=ddx(tan−1(x2+1))=11+(x2+1)2⋅2x\frac{dy}{dx} = \frac{d}{dx}\left(\tan^{-1}(x^2 + 1)\right) = \frac{1}{1 + (x^2 + 1)^2} \cdot 2xdxdy=dxd(tan−1(x2+1))=1+(x2+1)21⋅2x
Step 4: Simplify the Expression
First, simplify the denominator: 1+(x2+1)2=1+(x4+2×2+1)=x4+2×2+21 + (x^2 + 1)^2 = 1 + (x^4 + 2x^2 + 1) = x^4 + 2x^2 + 21+(x2+1)2=1+(x4+2×2+1)=x4+2×2+2
Now, substitute this back into the expression for dydx\frac{dy}{dx}dxdy: dydx=2xx4+2×2+2\frac{dy}{dx} = \frac{2x}{x^4 + 2x^2 + 2}dxdy=x4+2×2+22x
Final Answer:
dydx=2xx4+2×2+2\frac{dy}{dx} = \frac{2x}{x^4 + 2x^2 + 2}dxdy=x4+2×2+22x
Explanation:
The derivative dydx\frac{dy}{dx}dxdy is found by applying the chain rule to the inverse tangent function. First, differentiate the outer function, tan−1(u)\tan^{-1}(u)tan−1(u), where u=x2+1u = x^2 + 1u=x2+1, then multiply by the derivative of the inner function u=x2+1u = x^2 + 1u=x2+1, which is 2x2x2x. The denominator 1+(x2+1)21 + (x^2 + 1)^21+(x2+1)2 simplifies to x4+2×2+2x^4 + 2x^2 + 2×4+2×2+2.
