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Solve dy dx = 2xy 1+x2. left side is a side can be integrated using substitution: Let u = 1 + x2, so du = 2x dx.
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#. lny=(x^x)lnx#. #. ln(lny)=ln{(x^x)lnx}=ln(x^x)+ln(lnx)#, #i.e., ln(lny)=xlnx+ln(lnx)#. sides w.r.t. #x#, #d/dx{ln(lny
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Differentiating x to power of 1) If y = x n, dy/dx = nx n 2) If y = kx n, dy/dx = nkx n-1 k is a constant- in words a to differentiate x to power of you bring power down to in front of x, and reduce power by one
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A cut for differentiation is using partial derivative (β/βx). you use partial derivative, you treat all variables, except one you are differentiating respect to, like a constant. For β/βx [2xy + y^2] = 2y. In case, y is treated as a constant. is β/βy [2xy
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