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Mathematics

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Answer

4x(x2+1)2\frac{4x}{(x^2 + 1)^2}

Step 1: Identify the function and the differentiation rule needed.

The given function is f(x)=x21x2+1f(x) = \frac{x^2 - 1}{x^2 + 1}. This is a quotient of two functions, so we will use the quotient rule: If f(x)=g(x)h(x)f(x) = \frac{g(x)}{h(x)}, then f(x)=g(x)h(x)g(x)h(x)(h(x))2f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{(h(x))^2}.

Step 2: Define g(x)g(x) and h(x)h(x) and find their derivatives.

Let g(x)=x21g(x) = x^2 - 1. Then g(x)=ddx(x21)=2xg'(x) = \frac{d}{dx}(x^2 - 1) = 2x.

Let h(x)=x2+1h(x) = x^2 + 1. Then h(x)=ddx(x2+1)=2xh'(x) = \frac{d}{dx}(x^2 + 1) = 2x.

Step 3: Apply the quotient rule.

Substitute g(x)g(x), h(x)h(x), g(x)g'(x), and h(x)h'(x) into the quotient rule formula:

f(x)=(2x)(x2+1)(x21)(2x)(x2+1)2f'(x) = \frac{(2x)(x^2 + 1) - (x^2 - 1)(2x)}{(x^2 + 1)^2}

Step 4: Simplify the numerator.

f(x)=2x(x2+1(x21))(x2+1)2f'(x) = \frac{2x(x^2 + 1 - (x^2 - 1))}{(x^2 + 1)^2} f(x)=2x(x2+1x2+1)(x2+1)2f'(x) = \frac{2x(x^2 + 1 - x^2 + 1)}{(x^2 + 1)^2} f(x)=2x(2)(x2+1)2f'(x) = \frac{2x(2)}{(x^2 + 1)^2} f(x)=4x(x2+1)2f'(x) = \frac{4x}{(x^2 + 1)^2}

Final answer: 4x(x2+1)2\boxed{\frac{4x}{(x^2 + 1)^2}}

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