Find the differential dy of the given function:
y=xcosx
would this just be dy=(cosx-xsinx)dx
correct answer! dy/dx = cosx-xsinx.
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Then type x*cos[x]
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You will see solution STEP-BY-STEP
To find the differential dy of the function y = xcosx, you can use the product rule of differentiation.
We have y = x * cosx, where u = x and v = cosx. Applying the product rule, we have:
dy = (u * dv) + (v * du)
Taking the derivative of u (x) with respect to x gives du = 1. And taking the derivative of v (cosx) with respect to x gives dv = -sinx.
Plugging these values into the formula for dy, we get:
dy = (x * (-sinx)) + (cosx * 1)
= -xsinx + cosx
Therefore, the differential dy of the function y = xcosx is dy = -xsinx + cosx.
Yes, you are correct!
To find the differential dy of the function y = xcosx, you can use the differential operator dx, which represents an infinitesimally small change in x.
First, apply the product rule for differentiating y = xcosx:
dy/dx = (d(x) * cosx) + (x * d(cosx))
Now, differentiate the individual parts:
- d(x) = dx (since x is the variable with respect to which you are differentiating)
- d(cosx) = -sinx * dx (using the chain rule for differentiating cosx)
Substituting these differentials back in, we get:
dy/dx = (dx * cosx) + (x * (-sinx * dx))
Simplifying:
dy/dx = cosx*dx - x*sinx*dx
Factoring out dx from both terms:
dy/dx = (cosx - x*sinx) * dx
Therefore, the differential dy of the given function y = xcosx is dy = (cosx - x*sinx) * dx.