Find the 2nd Derivative f(x) = square root of x^2+1

Problem

ƒ(x)=√(,x2+1)

Solution

1. Rewrite the function using a fractional exponent to prepare for differentiation.

ƒ(x)=(x2+1)(1/2)

2. Apply the power rule and chain rule to find the first derivative ƒ(x)′

ƒ(x)′=1/2*(x2+1)(−1/2)⋅(d(x2)+1)/d(x)

3. Differentiate the inner function and simplify the expression.

ƒ(x)′=1/2*(x2+1)(−1/2)⋅2*x

ƒ(x)′=x*(x2+1)(−1/2)

4. Apply the product rule to find the second derivative ƒ(x)″ where the rule is (d(u)*v)/d(x)=u′*v+u*v′

ƒ(x)″=d(x)/d(x)*(x2+1)(−1/2)+x⋅d(x2+1)/d(x)

5. Differentiate the terms using the power rule and chain rule again.

ƒ(x)″=1⋅(x2+1)(−1/2)+x⋅(−1/2*(x2+1)(−3/2)⋅2*x)

6. Simplify the second term by multiplying the factors.

ƒ(x)″=(x2+1)(−1/2)−x2*(x2+1)(−3/2)

7. Factor out the common term (x2+1)(−3/2) to combine the expression.

ƒ(x)″=(x2+1)(−3/2)⋅((x2+1)1−x2)

8. Simplify the expression inside the parentheses.

ƒ(x)″=(x2+1)(−3/2)⋅1

ƒ(x)″=1/((x2+1)(3/2))

Final Answer

ƒ(x)″=1/((x2+1)√(,x2+1))

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