Suppose $u$ and $v$ are functions of $x$ that are differentiable at $x=0$ and that $u(0)=-3, u^{\prime}(0)=5$. $v(0)=3$, and $v^{\prime}(0)=4$. Find the values of the following derivatives at $x=0$.
a. $\frac{ d }{ dx }$ (uv)
b. $\frac{d}{d x}\left(\frac{u}{v}\right)$
c. $\frac{ d }{ dx }\left(\frac{ v }{ u }\right)$
d. $\frac{d}{d x}(-7 v-7 u)$
Answer (1)
Apply the product rule to find d x d ( uv ) and evaluate at x = 0 : 3 .
Apply the quotient rule to find d x d ( v u ) and evaluate at x = 0 : 3 .
Apply the quotient rule to find d x d ( u v ) and evaluate at x = 0 : − 3 .
Apply the constant multiple and sum/difference rules to find d x d ( − 7 v − 7 u ) and evaluate at x = 0 : $\boxed{-63}.
Explanation
Problem Analysis We are given two differentiable functions, u and v , and their values and derivatives at x = 0 . We need to find the derivatives of several expressions involving u and v , and evaluate them at x = 0 .
Applying the Product Rule a. We need to find d x d ( uv ) at x = 0 . Using the product rule, we have d x d ( uv ) = u ′ v + u v ′ At x = 0 , this becomes u ′ ( 0 ) v ( 0 ) + u ( 0 ) v ′ ( 0 ) = ( 5 ) ( 3 ) + ( − 3 ) ( 4 ) = 15 − 12 = 3
Applying the Quotient Rule (1) b. We need to find d x d ( v u ) at x = 0 . Using the quotient rule, we have d x d ( v u ) = v 2 u ′ v − u v ′ At x = 0 , this becomes [ v ( 0 ) ] 2 u ′ ( 0 ) v ( 0 ) − u ( 0 ) v ′ ( 0 ) = ( 3 ) 2 ( 5 ) ( 3 ) − ( − 3 ) ( 4 ) = 9 15 + 12 = 9 27 = 3
Applying the Quotient Rule (2) c. We need to find d x d ( u v ) at x = 0 . Using the quotient rule, we have d x d ( u v ) = u 2 v ′ u − v u ′ At x = 0 , this becomes [ u ( 0 ) ] 2 v ′ ( 0 ) u ( 0 ) − v ( 0 ) u ′ ( 0 ) = ( − 3 ) 2 ( 4 ) ( − 3 ) − ( 3 ) ( 5 ) = 9 − 12 − 15 = 9 − 27 = − 3
Applying Constant Multiple and Sum/Difference Rules d. We need to find d x d ( − 7 v − 7 u ) at x = 0 . Using the constant multiple rule and the sum/difference rule, we have d x d ( − 7 v − 7 u ) = − 7 v ′ − 7 u ′ = − 7 ( v ′ + u ′ ) At x = 0 , this becomes − 7 [ v ′ ( 0 ) + u ′ ( 0 )] = − 7 ( 4 + 5 ) = − 7 ( 9 ) = − 63
Final Answer Therefore, the values of the derivatives at x = 0 are: a. d x d ( uv ) = 3 b. d x d ( v u ) = 3 c. d x d ( u v ) = − 3 d. d x d ( − 7 v − 7 u ) = − 63
Examples
Understanding derivatives of combined functions is crucial in physics. For instance, if u ( t ) represents the velocity of an object and v ( t ) represents its mass, then the derivative of their product, d t d ( u ( t ) v ( t )) , gives the rate of change of the object's momentum. Similarly, in economics, if u ( x ) is the cost function and v ( x ) is the production level, then d x d ( v ( x ) u ( x ) ) represents the rate of change of the average cost of production.
