211k views
1 vote
Let y=(x²+3)⁵. Find the differential y when x=4 and x=0.1.

1 Answer

3 votes

Final answer:

The differentials of the function y=(x²+3)⁵ for x=4 and x=0.1 are found by first calculating the derivative dy/dx and then substituting the x values. The differential at x=4 is 5,226,480 and at x=0.1 is 82.

Step-by-step explanation:

The student's question is about finding the differential y for the given function y=(x²+3)⁵ when x = 4 and x=0.1. To find these differentials, we first need to compute the derivative of the function with respect to x, which is dy/dx. Applying the chain rule, the derivative of y with respect to x is dy/dx = 5(x²+3)⁴(2x). We can substitute x=4 and x=0.1 into the derived expression to calculate the differential at these points.

When x=4:
dy/dx = 5(4²+3)⁴(2×4) = 5(19)⁴(8) = 5(130,321)(8) = 5,226,480

When x=0.1:
dy/dx = 5(0.1²+3)⁴(2×0.1) = 5(3.01)⁴(0.2) = 5(82.0)(0.2) = 82

User Arturomp
by
7.2k points