If \( y = e^{{2}\log_e t} \) and \( x = \log_3(e^{t^2}) \), then \( \frac{dy}{dx} \) is equal to:
\( \frac{\log_e 3}{4t\sqrt{t}} \)
\({2t^2} \)
Step 1: Simplify $y$.
The given equation is:
$y = e^{\frac{1}{2} \log_e t}$.
Using logarithmic properties:
$y = t^{\frac{1}{2}} = \sqrt{t}$.
Step 2: Simplify $x$.
The given equation is:
$x = \log_3 (e^t)$.
Using $\log_a (b^c) = c \cdot \log_a b$:
$x = t \cdot \log_3 e$.
Step 3: Differentiate $y$ with respect to $t$.
$\frac{dy}{dt} = \frac{d}{dt} (t^{\frac{1}{2}}) = \frac{1}{2} t^{-\frac{1}{2}} = \frac{1}{2\sqrt{t}}$.
Step 4: Differentiate $x$ with respect to $t$.
$\frac{dx}{dt} = \frac{d}{dt} (t \cdot \log_3 e) = \log_3 e$.
Step 5: Compute $\frac{dy}{dx}$.
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{1}{2\sqrt{t}}}{\log_3 e}$.
Simplify:
$\frac{dy}{dx} = \frac{1}{2\sqrt{t} \cdot \log_3 e}$.
Using the property $\log_3 e = \frac{1}{\log_e 3}$, rewrite:
$\frac{dy}{dx} = \frac{\log_e 3}{2\sqrt{t}}$.
Final Answer:
$\frac{\log_e 3}{2\sqrt{t}}$
Let the function, \(f(x)\) = \(\begin{cases} -3ax^2 - 2, & x < 1 \\a^2 + bx, & x \geq 1 \end{cases}\) Be differentiable for all \( x \in \mathbb{R} \), where \( a > 1 \), \( b \in \mathbb{R} \). If the area of the region enclosed by \( y = f(x) \) and the line \( y = -20 \) is \( \alpha + \beta\sqrt{3} \), where \( \alpha, \beta \in \mathbb{Z} \), then the value of \( \alpha + \beta \) is:
List-I (Words) | List-II (Definitions) |
(A) Theocracy | (I) One who keeps drugs for sale and puts up prescriptions |
(B) Megalomania | (II) One who collects and studies objects or artistic works from the distant past |
(C) Apothecary | (III) A government by divine guidance or religious leaders |
(D) Antiquarian | (IV) A morbid delusion of one’s power, importance or godliness |