Question

1)y=(a^2)/2*ln(x+sqrt(x^2-a^2)-(b^2)/2 2)y=sqrt(x)*(e^x-1)*ln*x

Answer 1

$1)\; \; y=\frac{a^2}{2}\cdot ln(x+\sqrt{x^2-a^2})-\frac{b^2}{2}\\\\(lnu)'=\frac{1}{u}\cdot u'\; \; ,\; \; u=x+\sqrt{x^2-a^2}\\\\y'=\frac{a^2}{2}\cdot \frac{1}{x+\sqrt{x^2-a^2}}\cdot \Big (1+\frac{2x}{2\sqrt{x^2-a^2}}\Big )=\frac{a^2}{2}\cdot \frac{1}{x+\sqrt{x^2-a^2}}\cdot \frac{\sqrt{x^2-a^2}+x}{\sqrt{x^2-a^2}}=\\\\=\frac{a^2}{2\cdot \sqrt{x^2-a^2}}$

$2)\; \; y=\sqrt{x}\cdot (e^{x}-1)\cdot lnx\\\\(uvw)'=u'vw+uv'w+uvw'\\\\y'=\frac{1}{2\sqrt{x}}\cdot (e^{x}-1)\cdot lnx+\sqrt{x}\cdot e^{x}\cdot lnx+\sqrt{x}\cdot (e^{x}-1)\cdot \frac{1}{x}$

Answer 2

It seems like you have provided two equations. I'll try to clarify and solve them one by one:

1. $y = \frac{a^2}{2} \ln(x + \sqrt{x^2 - a^2}) - \frac{b^2}{2}$

2. $y = \sqrt{x} \cdot (e^x - 1) \ln x$

Let's tackle them individually:

1. $y = \frac{a^2}{2} \ln(x + \sqrt{x^2 - a^2}) - \frac{b^2}{2}$

This equation is not explicitly solvable for $x$ in terms of $y$ due to the presence of the natural logarithm. However, we can still analyze its behavior.

• Domain: For the expression inside the logarithm to be real, $x^2 - a^2$ must be greater than or equal to zero. This gives us $x \geq a$ or $x \leq -a$. However, the term $\sqrt{x^2 - a^2}$ should always be positive, so the valid domain for this equation is $x \geq a$.

• Asymptotes: As $x$ approaches $a$ from the right side, the term inside the logarithm approaches zero, so the function approaches negative infinity. As $x$ becomes very large, the logarithm term dominates the expression, and the function grows toward positive infinity.

• Intercepts: To find the y-intercept, we set $x = 0$ and calculate $y$. The expression becomes $y = \frac{a^2}{2} \ln(\sqrt{-a^2}) - \frac{b^2}{2}$. However, since the expression inside the logarithm is negative, this equation is not defined for real numbers, and there is no y-intercept.

2. $y = \sqrt{x} \cdot (e^x - 1) \ln x$

To solve this equation, we can set $y$ as a known value and try to solve for $x$. However, it is important to note that this equation might not have a closed-form solution for $x$ due to the presence of both a logarithm and an exponential term.

• Domain: For this equation to be valid, $x$ must be greater than zero because of the square root and the natural logarithm terms.

• Asymptotes: As $x$ approaches zero, the function approaches zero. As $x$ becomes very large, the exponential term dominates the expression, and the function grows rapidly.

• Intercepts: To find the x-intercept, we set $y = 0$ and solve for $x$. However, this equation is a bit complex, and a closed-form solution might not be straightforward to obtain.

In conclusion, while I've provided an analysis of both equations, please be aware that the first equation might not have an explicit solution for $x$ in terms of $y$. The second equation might be challenging to solve for specific values of $y$ without numerical methods. If you have specific values of $a$ and $b$ for the first equation or $y$ for the second equation, I can help you with numerical approximations.