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, well, well, if it isn't my favorite trigonometric function, tan! Let's have some fun with differentials, shall we?
When x = 1 and dx = 0.3, we want to find dy. So, let's first find the derivative of y with respect to x:
dy/dx = sec^2(2x+5) * 2
Now, let's substitute our values of x and dx:
dy = (sec^2(2(1)+5) * 2) * 0.3
And now it's time for the punchline!
dy ≈ "I tan't give you an exact answer, but I can approximate it as approximately 1.761 times dx."
Now, onto the second part!
When x = 1 and dx = 0.6, we'll go through the same process. Finding the derivative:
dy/dx = sec^2(2x+5) * 2
Substituting the values:
dy = (sec^2(2(1)+5) * 2) * 0.6
And here comes the punchline part deux!
dy ≈ "Now, this one's a bit bigger! Let's approximate it as approximately 3.522 times dx."
Remember, these are just approximate values, so take them with a grain of humor!
To find the differential dy, we can use the formula:
dy = f'(x) * dx
where f'(x) is the derivative of y with respect to x and dx is the given change in x.
First, let's find the derivative of y = tan(2x + 5) with respect to x.
Using the chain rule, we have:
dy/dx = sec^2(2x + 5) * 2
Now, let's calculate the differential dy when x = 1 and dx = 0.3.
dy = f'(1) * dx
= sec^2(2(1) + 5) * 2 * 0.3
= sec^2(7) * 0.6
For x = 1 and dx = 0.3, the value of dy is sec^2(7) * 0.6.
Continuing to find the differential dy when x = 1 and dx = 0.6:
dy = f'(1) * dx
= sec^2(2(1) + 5) * 2 * 0.6
= sec^2(7) * 1.2
For x = 1 and dx = 0.6, the value of dy is sec^2(7) * 1.2.
To find the differential dy when x = 1 and dx = 0.3, we can use the derivative of the given function y = tan(2x + 5) with respect to x.
Step 1: Find the derivative dy/dx of y = tan(2x + 5).
The derivative of y = tan(2x + 5) can be found using the chain rule.
dy/dx = sec^2(2x + 5) * (d/dx) (2x + 5)
dy/dx = sec^2(2x + 5) * 2
Step 2: Substitute x = 1 into the derivative expression.
dy/dx = sec^2(2(1) + 5) * 2
dy/dx = sec^2(7) * 2
Step 3: Find the differential dy using the derivative and the given value of dx.
dy = (dy/dx) * dx
dy = (sec^2(7) * 2) * 0.3
dy ≈ 2.159 * 0.3
dy ≈ 0.6477
Therefore, when x = 1 and dx = 0.3, the differential dy is approximately 0.6477.
To find the differential dy when x = 1 and dx = 0.6, follow the same steps:
Step 1: Find the derivative dy/dx of y = tan(2x + 5).
dy/dx = sec^2(2x + 5) * 2
Step 2: Substitute x = 1 into the derivative expression.
dy/dx = sec^2(2(1) + 5) * 2
dy/dx = sec^2(7) * 2
Step 3: Find the differential dy using the derivative and the given value of dx.
dy = (dy/dx) * dx
dy = (sec^2(7) * 2) * 0.6
dy ≈ 2.159 * 0.6
dy ≈ 1.2954
Therefore, when x = 1 and dx = 0.6, the differential dy is approximately 1.2954.