Find x'_y for: (a) y = ln x + e^x; (b) y = sqrt(sin x+

Find $x'_{y}$ for: (a) $y = \ln x + e^{x}$; (b) $y = \sqrt{\sin x+\cos x}$; (c) $y = \arcsin x$.

Step 1: Part (a): $y = \ln x + e^{x}$

Differentiate with respect to $x$, then invert: \[y'_{x} = \frac1x + e^{x} = \frac{1+xe^{x}}{x} \;\Longrightarrow\; x'_{y} = \frac{x}{1+xe^{x}}.\]

\[\text{(a)}\qquad x'_{y} = \dfrac{x}{1+xe^{x}}, \qquad x \gt 0\]

On $x \gt 0$ both terms of $y'_{x}$ are positive, so the inverse exists.

Step 2: Part (b): $y = \sqrt{\sin x+\cos x}$

Chain rule first: $y'_{x} = \frac{\cos x-\sin x}{2\sqrt{\sin x+\cos x}}$.

Inverting, and recognising the radical as $y$ itself: $x'_{y} = \frac{2\sqrt{\sin x+\cos x}}{\cos x-\sin x} = \frac{2y}{\cos x-\sin x}$.

\[\text{(b)}\qquad x'_{y} = \dfrac{2\sqrt{\sin x+\cos x}}{\cos x-\sin x} = \dfrac{2y}{\cos x-\sin x}\]

This fails where $\cos x = \sin x$, i.e. $x = \tfrac{\pi}{4}+k\pi$ — exactly the turning points of $\sin x+\cos x$, where $y$ has a horizontal tangent and the inverse a vertical one.

Step 3: Part (c): $y = \arcsin x$

\[ y'_{x} = \frac{1}{\sqrt{1-x^{2}}} \;\Longrightarrow\; x'_{y} = \sqrt{1-x^{2}}. \]

Since $y = \arcsin x$ means $x = \sin y$, this can be written purely in $y$: $x'_{y} = \sqrt{1-\sin^{2}y} = \cos y$.

\[\text{(c)}\qquad x'_{y} = \sqrt{1-x^{2}} = \cos y\]

Step 4: Verification

\[\text{(c) The inverse of } y = \arcsin x \text{ is } x = \sin y, \text{ whose derivative is } \cos y \;\;\checkmark\]

\[\text{(a) } x = 1:\quad y'_{x} = 1+e \approx 3.71828,\quad x'_{y} = \frac{1}{1+e} \approx 0.26894 \;\;\checkmark\]

\[\text{(b) } x = 0:\quad y = 1,\ y'_{x} = \tfrac12,\quad x'_{y} = 2 = \frac{2\cdot 1}{1-0} \;\;\checkmark\]