Let y=(x^2+3)^5. Find the differential dy when x=5 and dx=.5 Find the differential dy when x=5 and dx=.01

Question

Let y=(x^2+3)^5.

Find the differential dy when x=5 and dx=.5

Find the differential dy when x=5 and dx=.01

Answer 1

Let y = f(x) = (x² + 3)⁵

Then dy = f'(x)dx = 5(x² + 3)⁴(2x)dx = 10x(x² + 3)⁴dx

When x = 5 and dx = 0.5, we get dy = 50(28)⁴(0.5) = 15,366,400

Do the other problem in the same manner