57. Given that [tex]$y=3 e^{2 x}$[/tex], find [tex]$y^{\prime \prime \prime}(0)$[/tex].
58. Find [tex]$\frac{d y}{d x}$[/tex] if [tex]$y=\ln (\sin x)$[/tex]. A. [tex]$\cos x$[/tex] B. [tex]$\operatorname{cosec} x$[/tex] C.
59. If [tex]$y=\pi^3$[/tex]. find [tex]$\frac{d^2 y}{d x^2}$[/tex]. A. [tex]$6 \pi$[/tex] B. [tex]$\pi$[/tex] C. 0 D. does not exist
60. If [tex]$y=(x-1)^2$[/tex], find [tex]$\frac{d y}{d x}$[/tex]. A. [tex]$2(x-1)^3$[/tex] B. [tex]$2 x-2$[/tex]
61. If [tex]$y=x \ln x-x$[/tex], find [tex]$\frac{d y}{d x}$[/tex]. A. [tex]$x^2 \ln x$[/tex] B. [tex]$\ln x-1$[/tex]
62. If [tex]$y(t)=3 t^3+2 t^2-7 t+3$[/tex], find [tex]$\frac{d y}{d t}$[/tex] at [tex]$t=-1$[/tex].
63. If [tex]$y^2+x y-x=0$[/tex], find [tex]$\frac{d y}{d x}$[/tex] at [tex]$(0,2)$[/tex]. A. [tex]$\frac{1}{2}$[/tex] B. [tex]$\frac{2}{3}$[/tex]
64. Find value of [tex]$y^{\prime}$[/tex] if [tex]$x=3 t, y=t^2-4$[/tex] at [tex]$t=3$[/tex].
Answer (2)
The third derivative of y = 3 e 2 x evaluated at x = 0 is 24, and the derivative of y = ln ( sin x ) is cot x . Additionally, the second derivative of y = π 3 is 0, the derivative of y = ( x − 1 ) 2 is 2 x − 2 , and the derivative of y = x ln x − x simplifies to ln x .
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Calculate the third derivative of y = 3 e 2 x and evaluate at x = 0 , resulting in y ′′′ ( 0 ) = 24 .
Find the derivative of y = ln ( sin x ) , which simplifies to d x d y = cot x .
Determine the second derivative of y = π 3 , yielding d x 2 d 2 y = 0 .
Compute the derivative of y = ( x − 1 ) 2 , giving d x d y = 2 x − 2 .
Calculate the derivative of y = x ln x − x , which results in d x d y = ln x .
Evaluate the derivative of y ( t ) = 3 t 3 + 2 t 2 − 7 t + 3 at t = − 1 , obtaining d t d y ∣ t = − 1 = − 2 .
Use implicit differentiation to find d x d y for y 2 + x y − x = 0 at ( 0 , 2 ) , resulting in d x d y ∣ ( 0 , 2 ) = − 4 1 .
Calculate d x d y for x = 3 t and y = t 2 − 4 at t = 3 , which gives d x d y ∣ t = 3 = 2 .
24 , cot x , 0 , 2 x − 2 , ln x , − 2 , − 4 1 , 2
Explanation
Introduction We are given several calculus problems to solve. Let's address them one by one.
Question 57 Solution Question 57: Given y = 3 e 2 x , we need to find y ′′′ ( 0 ) .
First, find the first derivative: y ′ = d x d ( 3 e 2 x ) = 3 ⋅ 2 e 2 x = 6 e 2 x
Next, find the second derivative: y ′′ = d x d ( 6 e 2 x ) = 6 ⋅ 2 e 2 x = 12 e 2 x
Then, find the third derivative: y ′′′ = d x d ( 12 e 2 x ) = 12 ⋅ 2 e 2 x = 24 e 2 x
Finally, evaluate y ′′′ ( 0 ) :
y ′′′ ( 0 ) = 24 e 2 ( 0 ) = 24 e 0 = 24 ⋅ 1 = 24
Question 58 Solution Question 58: Find d x d y if y = ln ( sin x ) .
Using the chain rule: d x d y = sin x 1 ⋅ d x d ( sin x ) = sin x 1 ⋅ cos x = sin x cos x = cot x
Question 59 Solution Question 59: If y = π 3 , find d x 2 d 2 y .
Since y = π 3 is a constant, its first derivative is 0: d x d y = 0
Therefore, the second derivative is also 0: d x 2 d 2 y = 0
Question 60 Solution Question 60: If y = ( x − 1 ) 2 , find d x d y .
Using the power rule and chain rule: d x d y = 2 ( x − 1 ) ⋅ d x d ( x − 1 ) = 2 ( x − 1 ) ⋅ 1 = 2 ( x − 1 ) = 2 x − 2
Question 61 Solution Question 61: If y = x ln x − x , find d x d y .
Using the product rule and power rule: d x d y = d x d ( x ln x ) − d x d ( x ) = ( 1 ⋅ ln x + x ⋅ x 1 ) − 1 = ln x + 1 − 1 = ln x
Question 62 Solution Question 62: If y ( t ) = 3 t 3 + 2 t 2 − 7 t + 3 , find d t d y at t = − 1 .
First, find the derivative: d t d y = 9 t 2 + 4 t − 7
Now, evaluate at t = − 1 :
d t d y ∣ t = − 1 = 9 ( − 1 ) 2 + 4 ( − 1 ) − 7 = 9 − 4 − 7 = − 2
Question 63 Solution Question 63: If y 2 + x y − x = 0 , find d x d y at ( 0 , 2 ) .
Using implicit differentiation: 2 y d x d y + x d x d y + y − 1 = 0
d x d y ( 2 y + x ) = 1 − y
d x d y = 2 y + x 1 − y
At ( 0 , 2 ) :
d x d y = 2 ( 2 ) + 0 1 − 2 = 4 − 1
Question 64 Solution Question 64: Find the value of y ′ if x = 3 t , y = t 2 − 4 at t = 3 .
d t d x = 3 d t d y = 2 t
d x d y = d t d x d t d y = 3 2 t
At t = 3 :
d x d y = 3 2 ( 3 ) = 2
Final Answers Here are the final answers to each question:
y ′′′ ( 0 ) = 24
d x d y = cot x
d x 2 d 2 y = 0
d x d y = 2 x − 2
d x d y = ln x
d t d y ∣ t = − 1 = − 2
d x d y ∣ ( 0 , 2 ) = − 4 1
d x d y ∣ t = 3 = 2
Examples
Understanding derivatives is crucial in physics, especially when analyzing motion. For instance, if you have a function describing the position of an object over time, the first derivative gives you the velocity, and the second derivative gives you the acceleration. These concepts are fundamental in understanding how objects move and interact with forces.
