66. If [tex]f(x)=\left(3+e^x\right)^2[/tex], then [tex]f^{\prime \prime}(0)[/tex].
A. 10
B. -1
67. Find [tex]y^{\prime}[/tex] if [tex]y=\left(\frac{x-1}{x+1}\right)^2[/tex] at [tex]x=0[/tex].
A. i)
B. -2
C.
68. The derivative [tex]\frac{d y}{d x}[/tex] of [tex]x^2-x y=0[/tex] at [tex](1,0)[/tex] is
69. If [tex]V=t^3+5 t[/tex], find [tex]\dot{V}[/tex] at [tex]t=0[/tex].
A. -5
B. 0
70. Find the derivative of [tex]x^2+y^2=5[/tex].
A. [tex]\frac{x}{y}[/tex]
B.
71. The derivative of [tex]y=x \ln x[/tex] at [tex]x=0[/tex] is A.
72. Find [tex]\frac{d y}{d x}[/tex] if [tex]y=\cos 2 t[/tex] and [tex]x=\sin 2 t[/tex].
A. [tex]-\cot[/tex]
D. [tex]2 \cos 2 t[/tex].
73. Find [tex]y^{\prime \prime}(0)[/tex] if [tex]y=3 e^{2 x}+\sin x[/tex].
A. 3
B. -12
Answer (1)
Find the first derivative f ′ ( x ) of f ( x ) = ( 3 + e x ) 2 using the chain rule: f ′ ( x ) = 2 ( 3 + e x ) e x .
Find the second derivative f ′′ ( x ) using the product rule and chain rule: f ′′ ( x ) = 4 e 2 x + 6 e x .
Evaluate f ′′ ( 0 ) : f ′′ ( 0 ) = 4 e 2 ( 0 ) + 6 e 0 = 10 .
The final answer is 10 .
Explanation
Problem Analysis We are given the function f ( x ) = ( 3 + e x ) 2 and we need to find its second derivative evaluated at x = 0 , i.e., f ′′ ( 0 ) .
Finding the First Derivative First, we find the first derivative f ′ ( x ) using the chain rule. The chain rule states that if we have a composite function f ( g ( x )) , then its derivative is f ′ ( g ( x )) ⋅ g ′ ( x ) . In our case, f ( u ) = u 2 and u ( x ) = 3 + e x . So, f ′ ( u ) = 2 u and u ′ ( x ) = e x . Therefore, f ′ ( x ) = 2 ( 3 + e x ) e x .
Finding the Second Derivative Next, we find the second derivative f ′′ ( x ) using the product rule and the chain rule. The product rule states that if we have a function h ( x ) = u ( x ) v ( x ) , then its derivative is h ′ ( x ) = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) . In our case, u ( x ) = 2 ( 3 + e x ) and v ( x ) = e x . So, u ′ ( x ) = 2 e x and v ′ ( x ) = e x . Therefore, f ′′ ( x ) = 2 e x "." e x + 2 ( 3 + e x ) e x = 2 e 2 x + 6 e x + 2 e 2 x = 4 e 2 x + 6 e x .
Evaluating f''(0) Finally, we evaluate f ′′ ( 0 ) . We have f ′′ ( 0 ) = 4 e 2 ( 0 ) + 6 e 0 = 4 e 0 + 6 e 0 = 4 ( 1 ) + 6 ( 1 ) = 4 + 6 = 10 .
Final Answer Therefore, f ′′ ( 0 ) = 10 .
Examples
Understanding derivatives is crucial in physics, especially when dealing with motion. For example, if f ( x ) represents the position of an object at time x , then f ′ ( x ) gives the object's velocity, and f ′′ ( x ) gives the object's acceleration. Evaluating these at specific times, like x = 0 , helps us understand the object's instantaneous state of motion. This is fundamental in fields like aerospace engineering for trajectory calculations and control systems.
