$f(x) = \int_0^x\frac{1-t^2}{\sqrt{t^4+1}}dt$ find it's derivative and tangent where x = 0 **I am given this function:** $$f(x) = \int_0^x\frac{1-t^2}{\sqrt{t^4+1}}dt$$ I have to find it's derivative $f'(x)$ and I have to find the equation of it's tangent in the point $x = 0$. I'm a bit confused about this one. **I think it's derivative is:** $$\frac{1-x^2}{\sqrt{x^4+1}}$$ Is that right? I've tried finding it's tangent, where $x = 0$. I've found $k_t = 1$, $x_0 = 0$, $y_0 = f(0)$ and inserted this into the equation: $$y - y_0 = k_t(x - x_0)$$ $$y = x - f(0)$$ Did I do this correctly or did I entirely miss the point?

Your solution is correct. It is a common error to "overthink" the calculation of $f'$. One thing you left out is the evaluation of $f(0)$. What is it?