The equation of the tangent at $f(x)=\int_3^{\sqrt{x}}e^{-t^2}dt$.

Question

Find the equation of the tangent line to the graph of $$f(x)=\int_3^{\sqrt{x}}e^{-t^2}dt$$ at $x = 9.$

I am trying to evaluate that problem.

I have obtained $$f'(x)=e^{-t^2}\Bigg|_3^{\sqrt{x}}=e^{-x}-e^9\Bigg|_3^{\sqrt{x}}=0.$$

I have failed to obtain $f(x)$ as the integration is hard.

Answer 1

Let $F(t)$ be the antiderivative of $e^{-t^2}$, i.e. $F'(t) = e^{-t^2}$.

Then, $f(x) = F(x^{0.5}) - F(3)$.

So $f'(x) = F'(x^{0.5}) (0.5 x^{-0.5}) = 0.5 e^{-x} x^{-0.5}$

Answer 2

$f'(x)=e^{-x} \frac d {dx} \sqrt x =e^{-x}\frac 1 2 x ^{-1/2}$ so $f'(9)=e^{-9}\frac 1 6$. The equation is $y-0=e^{-9}\frac 1 6 (x-9)$