$x=\frac{1}{4}(t+e^{-t})$, $y=e^{-\frac{t}{2}}$, prove that $yy''(x)=(y'(x))^2\cdot \sqrt{1+(y'(x))^2}$: what am I doing wrong?

Question

$x=\frac{1}{4}(t+e^{-t})$, $y=e^{-\frac{t}{2}}$, prove that $yy''(x)=(y'(x))^2\cdot \sqrt{1+(y'(x))^2}$

$$yy''(x)=e^{-\frac{t}{2}}\cdot \frac{(e^{-t/2})'' (\frac{1}{4}(t+e^{-t})'-(\frac{1}{4}(t+e^{-t}))''(e^{-t/2})'}{((e^{-t/2})')^3}$$

Derivating and simplifying, I get $$-\frac{1}{2}(\sqrt{e^t}+\frac{1}{\sqrt{e^t}}).$$ Which is a negative value, and can't be equal to what the exercise requires.

I am very sorry I didn't type out my 'interim calculations': MathJax is time-comsuming, it's late, and there's still homework left.

What I need to know, was there an arithmetical mistake, or do I misunderstand something fundamentally? Did I understand the exercise incorrectly? Did I get the wrong somewhere?

Thank you.

Answer 1

My mistake, I have put $y$ into the denominator instead of $x$.