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Y = LNX + xtanx, find the derivative of Y
y=lnlnx+xtanx
y'=(1/lnx)*(lnx)'+tanx+x*sec ² x
=1/xlnx+tanx+xsec ² x
What is the derivative of y = xtanx-2secx
y '=(xtanx) '-(2secx) '
=tanx+x(tanx) '-2secx tanx
=tanx+xsec ² x-2secx tanx
Find the derivative of this, y = cos2x + xtanx
y=cos2x+xtanx
y'=(cos2x)'+(xtanx)'
=-sin2x*(2x)'+tanx+x(tanx)'
=-2sin2x+tanx+xsec^2x.
How to calculate the original function of derivative 8xy I still don't understand ∫x(x)=∫[x,1]8xydy=4x(1-x^2)
Separate variables
dy/dx=8xy
dy/y=8xdx
lny = 4x^2
Get the original function:
y=e^(4x^2) -------- ①
The verification is as follows: find the derivative about X for both sides in ①:
dy/dx=8xe^(4x^2)
And y '= 8xy = 8XE ^ (4x ^ 2) (y = e ^ (4x ^ 2) is directly substituted into 8xy)
The two are equal
Can the average rate of change of the function be calculated by derivative? Please specify, Note: it is the average rate of change, not the instantaneous speed
Derivative is the limit of average rate of change. Use the definition of average rate of change to calculate. Derivative is a special case of average rate of change
Function y = cos ² X-sin ² Minimum period of X?
π
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