$\frac{d y}{d x}$ if $y=x \ln x-x$.
A. $x^2 \ln x$
B. $\ln x-1$
C. $\ln x$
D. $x-\ln x$.
$\frac{d y}{d t}$ of $y(t)=3 t^3+2 t^2-7 t+3$ at $t=-1$.
A. -1
B. 1
C. -2
D. 2.
$\frac{d y}{d x}$ of $1^2+x y-x=0$ at $(0,2)$.
A. $\frac{1}{2}$
B. $\frac{2}{3}$
C. $\frac{1}{3}$
D. $-\frac{1}{4}$.
Find the value of $y^{\prime}$ if $x=3 t, y=t^2-4$ at $t=3$.
A. 2
B. $\frac{1}{2}$
C. -4
D. -2 .
Find $f^{\prime}(0)$ if $f(x)=\frac{x+1}{x^3-1}, x \neq 1$.
A. 2
B. 3
C. -2
D. -1
Answer (1)
Apply the product rule to find the derivative of x ln x , which is 1 + ln x .
Differentiate − x to get − 1 .
Combine the results to find d x d y = ( 1 + ln x ) − 1 = ln x .
The final answer is ln x .
Explanation
Problem Analysis We are given y = x ln x − x and asked to find d x d y . This requires us to use the product rule for differentiation.
Finding the Derivative Let's differentiate y = x ln x − x with respect to x . We have: d x d y = d x d ( x ln x ) − d x d ( x ) Using the product rule, d x d ( x ln x ) = x ⋅ d x d ( ln x ) + ln x ⋅ d x d ( x ) = x ⋅ x 1 + ln x ⋅ 1 = 1 + ln x A l so , \frac{d}{dx}(x) = 1 . T h ere f ore , d x d y = ( 1 + ln x ) − 1 = ln x $
Final Answer Comparing our result with the given options, we see that the correct answer is C.
Examples
Understanding derivatives is crucial in physics, especially when dealing with motion. For example, if the position of a particle is given by s ( t ) = t ln t − t , then the velocity of the particle at any time t is given by the derivative s ′ ( t ) = ln t . This allows us to determine how the particle's speed changes over time, which is essential for analyzing its motion.
