1. Find [tex]\frac{d y}{d x}[/tex] if [tex]$y=\cos ^2 x+\sin ^2 x$[/tex].
2. Find [tex]\frac{d y}{d x}[/tex] if [tex]$3+y^2=x^3-y$[/tex].
3. Given [tex]$y(x)=3 e^{2 x}+\sin x$[/tex], find [tex]$y^{\prime \prime}(0)$[/tex].
4. Given [tex]$y=6 x e^{-5 x}$[/tex], find [tex]$\frac{d y}{d x}$[/tex].
5. Find the derivative of [tex]$y=\frac{x^2-1}{2 x+1}$[/tex] at [tex]$x=0$[/tex]
Answer (2)
Find the derivative of y = cos 2 x + sin 2 x by recognizing it simplifies to y = 1 , thus d x d y = 0 .
Use implicit differentiation on 3 + y 2 = x 3 − y to find d x d y = 2 y + 1 3 x 2 .
Calculate the second derivative of y ( x ) = 3 e 2 x + sin x and evaluate at x = 0 , resulting in y ′′ ( 0 ) = 12 .
Apply the product rule to y = 6 x e − 5 x to get d x d y = 6 e − 5 x ( 1 − 5 x ) .
Use the quotient rule on y = 2 x + 1 x 2 − 1 and evaluate at x = 0 , giving y ′ ( 0 ) = 2 .
Explanation
Introduction We will solve each derivative problem step by step.
Question 1 Solution Question 1: Find d x d y if y = cos 2 x + sin 2 x . We know that cos 2 x + sin 2 x = 1 . Therefore, y = 1 . The derivative of a constant is 0. So, d x d y = 0 .
Question 2 Solution Question 2: Find d x d y if 3 + y 2 = x 3 − y . We will use implicit differentiation. Differentiating both sides with respect to x , we get 0 + 2 y d x d y = 3 x 2 − d x d y . Rearranging the terms, we have ( 2 y + 1 ) d x d y = 3 x 2 . Therefore, d x d y = 2 y + 1 3 x 2 . Note that the options provided in the question are slightly different, but the correct expression is 2 y + 1 3 x 2 .
Question 3 Solution Question 3: Given y ( x ) = 3 e 2 x + sin x , find y ′′ ( 0 ) . First, we find the first derivative: y ′ ( x ) = 3 ( 2 ) e 2 x + cos x = 6 e 2 x + cos x . Next, we find the second derivative: y ′′ ( x ) = 6 ( 2 ) e 2 x − sin x = 12 e 2 x − sin x . Now, we evaluate y ′′ ( 0 ) : y ′′ ( 0 ) = 12 e 2 ( 0 ) − sin ( 0 ) = 12 ( 1 ) − 0 = 12 .
Question 4 Solution Question 4: Given y = 6 x e − 5 x , find d x d y . We will use the product rule: d x d y = u ′ v + u v ′ , where u = 6 x and v = e − 5 x . Then, u ′ = 6 and v ′ = − 5 e − 5 x . So, d x d y = 6 e − 5 x + 6 x ( − 5 e − 5 x ) = 6 e − 5 x − 30 x e − 5 x = 6 e − 5 x ( 1 − 5 x ) .
Question 5 Solution Question 5: Find the derivative of y = 2 x + 1 x 2 − 1 at x = 0 . We will use the quotient rule: y ′ = v 2 u ′ v − u v ′ , where u = x 2 − 1 and v = 2 x + 1 . Then, u ′ = 2 x and v ′ = 2 . So, y ′ = ( 2 x + 1 ) 2 2 x ( 2 x + 1 ) − ( x 2 − 1 ) ( 2 ) = ( 2 x + 1 ) 2 4 x 2 + 2 x − 2 x 2 + 2 = ( 2 x + 1 ) 2 2 x 2 + 2 x + 2 . Now, we evaluate y ′ ( 0 ) : y ′ ( 0 ) = ( 2 ( 0 ) + 1 ) 2 2 ( 0 ) 2 + 2 ( 0 ) + 2 = 1 2 = 2 .
Final Answers Final Answers:
d x d y = 0
d x d y = 2 y + 1 3 x 2
y ′′ ( 0 ) = 12
d x d y = 6 e − 5 x ( 1 − 5 x )
y ′ ( 0 ) = 2
Examples
Derivatives are a fundamental tool in calculus with numerous real-world applications. For instance, in physics, derivatives are used to calculate velocity and acceleration. In economics, they help determine marginal cost and revenue. In engineering, derivatives are essential for optimizing designs and predicting system behavior. Understanding derivatives allows us to analyze rates of change and make informed decisions in various fields.
The derivatives for the problems are as follows: 1) d x d y = 0 , 2) d x d y = 2 y + 1 3 x 2 , 3) y ′′ ( 0 ) = 12 , 4) d x d y = 6 e − 5 x ( 1 − 5 x ) , 5) (y'(0) = 2.
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