Let y=tan(2x+5).
Find the differential dy when x=1 and dx=.3
Find the differential dy when x=1 and dx=.6
well, since
dy = 2sec^2(2x+5) dx
just plug in your numbers
I can't figure out the first one???
Steve
Read second post.
To find the differential dy when x=1 and dx=.3, we can use the derivative of the function y=tan(2x+5). The derivative of tan(x) is sec^2(x), so the derivative of y with respect to x is dy/dx = sec^2(2x+5).
Now, to find the differential dy, we can use the formula: dy = (dy/dx) * dx.
When x=1 and dx=.3, we substitute these values into the derivative expression:
dy/dx = sec^2(2x+5)
dy/dx = sec^2(2(1)+5)
dy/dx = sec^2(7)
Now, substitute dy/dx = sec^2(7) and dx = .3 into the differential formula:
dy = (sec^2(7)) * .3
Calculating the value using a scientific calculator or programming language, we find:
dy ≈ (2.853) * .3
dy ≈ 0.856
So, when x=1 and dx=.3, the differential dy is approximately 0.856.
To find the differential dy when x=1 and dx=.6, we follow the same procedure. First, calculate the derivative expression:
dy/dx = sec^2(2(1)+5)
dy/dx = sec^2(7)
Now, substitute dy/dx = sec^2(7) and dx = .6 into the differential formula:
dy = (sec^2(7)) * .6
Calculating the value:
dy = (2.853) * .6
dy ≈ 1.712
So, when x=1 and dx=.6, the differential dy is approximately 1.712.