Question

Differentiating a Logarithmic Function in Exercise, find the derivative of the function. See Examples 1, 2, 3, and 4.

y = ln x/x^2 + 1

Answer 1

`(d)/(dx)\left(\ln \left((x)/(x^2+1)\right)\right)=\left(\ln{\left((x)/(x^(2) + 1) \right)}\right)^(\prime )=(-x^2+1)/(x\left(x^2+1\right))`

Explanation:

To find the derivative of the function
`y(x)=\ln \left((x)/(x^2+1)\right)` you must:

Step 1. Rewrite the logarithm:

`\left(\ln{\left((x)/(x^(2) + 1) \right)}\right)^(\prime )=\left(ln(\left(x \right)) - \ln{\left(x^(2) + 1 \right)}\right)^(\prime )`

Step 2. The derivative of a sum is the sum of derivatives:

`\left(ln(\left(x \right)) - \ln{\left(x^(2) + 1 \right)}\right)^(\prime )}={\left(\left(ln(\left(x \right))\right)^(\prime ) - \left(\ln{\left(x^(2) + 1 \right)}\right)^(\prime )\right)`

Step 3. The derivative of natural logarithm is
`\left(ln(\left(x \right))\right)^(\prime )=(1)/(x)`

`{\left(ln(\left(x \right))\right)^(\prime )} - \left(\ln{\left(x^(2) + 1 \right)}\right)^(\prime )={(1)/(x)} - \left(\ln{\left(x^(2) + 1 \right)}\right)^(\prime )`

Step 4. The function
`\ln{\left(x^(2) + 1 \right)}` is the composition
`f\left(g\left(x\right)\right)` of two functions
`f\left(u\right)=ln(\left(u \right))` and
`u=g\left(x\right)=x^(2) + 1`

Step 5. Apply the chain rule
`\left(f\left(g\left(x\right)\right)\right)^(\prime )=(d)/(du)\left(f\left(u\right)\right) \cdot \left(g\left(x\right)\right)^(\prime )`

`-{\left(\ln{\left(x^(2) + 1 \right)}\right)^(\prime )} + (1)/(x)=- {(d)/(du)\left(ln(\left(u \right))\right) (d)/(dx)\left(x^(2) + 1\right)} + (1)/(x)\\\\- {(d)/(du)\left(ln(\left(u \right))\right)} (d)/(dx)\left(x^(2) + 1\right) + (1)/(x)=- {(1)/(u)} (d)/(dx)\left(x^(2) + 1\right) + (1)/(x)`

Return to the old variable:

`- \frac{1}{{u}} (d)/(dx)\left(x^(2) + 1\right) + (1)/(x)=- \frac{(d)/(dx)\left(x^(2) + 1\right)}{{\left(x^(2) + 1\right)}} + (1)/(x)`

The derivative of a sum is the sum of derivatives:

`- \frac{{(d)/(dx)\left(x^(2) + 1\right)}}{x^(2) + 1} + (1)/(x)=- \frac{{\left((d)/(dx)\left(1\right) + (d)/(dx)\left(x^(2)\right)\right)}}{x^(2) + 1} + (1)/(x)=(1)/(x^(3) + x) \left(x^(2) - x \left((d)/(dx)\left(1\right) + (d)/(dx)\left(x^(2)\right)\right) + 1\right)`

Step 6. Apply the power rule
`(d)/(dx)\left(x^(n)\right)=n\cdot x^(-1+n)`

`(1)/(x^(3) + x) \left(x^(2) - x \left({(d)/(dx)\left(x^(2)\right)} + (d)/(dx)\left(1\right)\right) + 1\right)=\\\\(1)/(x^(3) + x) \left(x^(2) - x \left({\left(2 x^(-1 + 2)\right)} + (d)/(dx)\left(1\right)\right) + 1\right)=\\\\(1)/(x^(3) + x) \left(- x^(2) - x (d)/(dx)\left(1\right) + 1\right)\\`

`(1)/(x^(3) + x) \left(- x^(2) - x {(d)/(dx)\left(1\right)} + 1\right)=\\\\(1)/(x^(3) + x) \left(- x^(2) - x {\left(0\right)} + 1\right)=\\\\(1 - x^(2))/(x \left(x^(2) + 1\right))`

Thus,
`(d)/(dx)\left(\ln \left((x)/(x^2+1)\right)\right)=\left(\ln{\left((x)/(x^(2) + 1) \right)}\right)^(\prime )=(-x^2+1)/(x\left(x^2+1\right))`