Finding the Value of a Derivative with Given Function and Derivative Values

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To find the derivative of the product of two differentiable functions u and v at x=1, the product rule is applied: d/dx(uv) = u*v' + v*u'. Given the values u(1)=2, u'(1)=-7, v(1)=7, and v'(1)=-2, the calculation becomes (2)(-2) + (7)(-7). This results in -4 - 49, yielding a total of -53. The solution confirms the correct application of the product rule and the evaluation at the specified point.
carlarae
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Homework Statement


Suppose u and v are differentiable functions of x. Use the given values of the functions and their derivatives to find the value of the indicated derivative.
u(1)=2, u'(1)=-7, v(1)=7,v'(1)=-2
d/dx (uv) at x =1


Homework Equations





The Attempt at a Solution

 
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Homework Equations


look for the equations you use when you have to differentiate a combination of two functions... one of them looks like your problem.
 
I get d/x(uv)=(2)(-2) + (7)(-7) = -53 but I'm not applying the 1 anywhere that I know of here. as in u(1), does anyone have an example that could help me?
 
carlarae said:
I get d/x(uv)=(2)(-2) + (7)(-7) = -53 but I'm not applying the 1 anywhere that I know of here. as in u(1), does anyone have an example that could help me?

In your answer (2)(2) + (7)(-7), where did the 2 come from? What about the -2? Where did you get the 7? Ditto for the -7.

RGV
 
That answer looks correct to me. You are applying the 1. The equation for the derivative of the product of two functions is u*v' + v*u'. In your case, you have u(1)*v'(1) + v(1)*u'(1) = (2)(-2) + (7)(-7) = -53. This is d/dx(uv) evaluated at x=1.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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