If t = e2x and y = loge(t2), then d2y/dx2 is :
0
We are given two equations relating the variables $t$, $x$, and $y$:
Our goal is to find the second derivative of $y$ with respect to $x$, which is denoted as $\frac{d^2y}{dx^2}$.
First, let's simplify the expression for $y$. We have $y = \log_e(t^2)$. Using the logarithm property $\log(a^b) = b \log(a)$, we can rewrite this as:
$\qquad y = 2 \log_e(t)$
Now, we substitute the given expression for $t$, which is $t = e^{2x}$, into the simplified equation for $y$:
$\qquad y = 2 \log_e(e^{2x})$
Using the property $\log_e(e^a) = a$, we can simplify this further:
$\qquad y = 2 \times (2x)$
$\qquad y = 4x$
So, the relationship between $y$ and $x$ is simply $y = 4x$. This makes finding the derivatives much easier.
Now, we find the first derivative of $y$ with respect to $x$. We need to differentiate $y = 4x$ with respect to $x$:
$\qquad \frac{dy}{dx} = \frac{d}{dx}(4x)$
Using the constant multiple rule and the power rule for differentiation ($\frac{d}{dx}(x) = 1$):
$\qquad \frac{dy}{dx} = 4 \times \frac{d}{dx}(x)$
$\qquad \frac{dy}{dx} = 4 \times 1$
$\qquad \frac{dy}{dx} = 4$
The first derivative $\frac{dy}{dx}$ is a constant value, 4.
Finally, we find the second derivative by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$. We need to differentiate the constant value 4 with respect to $x$:
$\qquad \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$
$\qquad \frac{d^2y}{dx^2} = \frac{d}{dx}(4)$
The derivative of any constant is always 0:
$\qquad \frac{d^2y}{dx^2} = 0$
Therefore, the second derivative of $y$ with respect to $x$ is 0.
Given $t = e^{2x}$ and $y = \log_e(t^2)$, we found that $y$ simplifies to $y = 4x$. The first derivative $\frac{dy}{dx}$ is 4, and the second derivative $\frac{d^2y}{dx^2}$ is 0.
| Step | Calculation | Result |
|---|---|---|
| Start with $y$ in terms of $t$ | $y = \log_e(t^2)$ | $y = 2 \log_e(t)$ |
| Substitute $t$ in terms of $x$ | $y = 2 \log_e(e^{2x})$ | $y = 4x$ |
| Calculate First Derivative $\frac{dy}{dx}$ | $\frac{d}{dx}(4x)$ | $4$ |
| Calculate Second Derivative $\frac{d^2y}{dx^2}$ | $\frac{d}{dx}(4)$ | $0$ |
| Concept | Rule/Definition | Example |
|---|---|---|
| Logarithm Property | $\log_b(a^c) = c \log_b(a)$ | $\log_e(t^2) = 2 \log_e(t)$ |
| Logarithm Property | $\log_b(b^c) = c$ | $\log_e(e^{2x}) = 2x$ |
| Derivative of $cx$ | $\frac{d}{dx}(cx) = c$ | $\frac{d}{dx}(4x) = 4$ |
| Derivative of a Constant | $\frac{d}{dx}(c) = 0$ | $\frac{d}{dx}(4) = 0$ |
The second derivative $\frac{d^2y}{dx^2}$ represents the rate of change of the first derivative $\frac{dy}{dx}$. While the first derivative tells us about the slope of a function at a point (how fast $y$ is changing with respect to $x$), the second derivative tells us about the concavity of the function (how the slope is changing). If the second derivative is positive, the curve is concave up. If it's negative, the curve is concave down. If the second derivative is zero, as in this case, it means the rate of change of the slope is zero. For the function $y=4x$, which is a straight line, the slope is constant (always 4), so its rate of change is indeed 0.
This problem demonstrates how simplifying the function expression first can significantly ease the differentiation process. Recognizing the direct relationship between $y$ and $x$ after substitution and applying logarithm properties was key.
The degree of the differential equation
\[\begin{equation*} \left[ 1- \frac{dy}{dx}\right]^{3/2} = k\frac{d^2y}{dx^2} \end{equation*}\]is :
Match List-I with List-II:
| List-I | List-II |
|---|---|
| (A) Integrating factor of \( xdy - (y + 2x^2)dx = 0 \) | (I) \( \frac{1}{x} \) |
| (B) Integrating factor of \( (2x^2 - 3y)dx = xdy \) | (II) \( x \) |
| (C) Integrating factor of \( (2y + 3x^2)dx + xdy = 0 \) | (III) \( x^2 \) |
| (D) Integrating factor of \( 2xdy + (3x^3 + 2y)dx = 0 \) | (IV) \( x^3 \) |
Choose the correct answer from the options given below:
If \( f(x) = 2 \left( \tan^{-1}(e^x) - \frac{\pi}{4} \right) \), then \( f(x) \) is:
Degree of the differential equation \( \frac{d^2y}{dx^2} + 3 \left( \frac{dy}{dx} \right)^{\frac{1}{2}} = y^2 + e^x \) is:
General solution of the differential equation \( \frac{2y dx - 3x dy}{y} = 0 \) is (c is an arbitrary constant):