Question

How do you find the derivative of `ln(x^2+1)`?

Answer 1

`frac{d}{dx}(y)=frac{2x}{x^2+1}`

Explanation:

Given equation is `y=ln(x^2+1)`

One fundamental equation of derivative calculus is
`frac{d}{dx}(y)=1/x` when `y=lnx`

Another important aspect of derivative calculus is the chain rule, that is if `y=f(t)` and `t=g(x)`, then
`dy/dx=dy/dt*dt/dx`

Taking `t=x^2+1`, from chain rule

`dy/dx=\frac{d}{dt}(lnt)*dt/dx`
`\implies dy/dx=1/t*fracddx(x^2+1)`

Substitute for whatever `t` remains in the equation as `x^2+1` and you'll get the answer.