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Category: Differentiation

one-of-the-conditions-of-the-inflection-point-is-inflection-tangent-what-is-inflection-tangent-

Question Number 91460 by  M±th+et+s last updated on 30/Apr/20 $${one}\:{of}\:{the}\:{conditions}\:{of}\:{the}\:{inflection} \\ $$$${point}\:{is}\:{inflection}\:{tangent}. \\ $$$${what}\:{is}\:{inflection}\:{tangent}? \\ $$ Answered by MJS last updated on 01/May/20 $$\mathrm{it}'\mathrm{s}\:\mathrm{the}\:\mathrm{tangent}\:\mathrm{in}\:\mathrm{the}\:\mathrm{inflection}\:\mathrm{point}\:\mathrm{which} \\…

0-1-ln-1-x-2-1-x-2-dx-proof-0-1-ln-1-x-1-x-2-dx-pi-8-ln-2-I-0-1-ln-1-x-1-x-2-dx-x-tan

Question Number 156864 by mnjuly1970 last updated on 16/Oct/21 $$ \\ $$$$\:\:\:\:\phi\::=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}\:\left(\mathrm{1}−{x}^{\:\mathrm{2}} \right)}{\mathrm{1}+\:{x}^{\:\mathrm{2}} }\:{dx}\:= \\ $$$$\:\:{proof}\:: \\ $$$$\:\:\:\:\phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}\left(\mathrm{1}−{x}\:\right)}{\mathrm{1}+{x}^{\:\mathrm{2}} }{dx}\:+\:\frac{\pi}{\mathrm{8}}{ln}\left(\mathrm{2}\right) \\ $$$$\:\:\:\:….\:\mathrm{I}=\:\int_{\mathrm{0}}…

Solve-the-differential-equation-d-2-y-dx-2-x-2-dy-dx-xy-x-

Question Number 91329 by niroj last updated on 29/Apr/20 $$\:\mathrm{S}\boldsymbol{\mathrm{olve}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{differential}}\:\boldsymbol{\mathrm{equation}}: \\ $$$$\:\:\:\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2}} }\:−\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \frac{\boldsymbol{\mathrm{dy}}}{\boldsymbol{\mathrm{dx}}}+\boldsymbol{\mathrm{xy}}\:=\:\boldsymbol{\mathrm{x}} \\ $$ Commented by MWSuSon last updated on 30/Apr/20 $${do}\:{you}\:{need}\:{a}\:{specific}\:{method}?…

If-x-z-tan-1-yz-and-z-z-x-y-find-z-x-z-y-

Question Number 156824 by Tawa11 last updated on 15/Oct/21 $$\mathrm{If}\:\:\:\:\:\:\mathrm{x}\:\:\:−\:\:\:\mathrm{z}\:\:\:\:=\:\:\:\:\mathrm{tan}^{−\:\mathrm{1}} \left(\mathrm{yz}\right)\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\mathrm{z}\:\:\:=\:\:\:\mathrm{z}\left(\mathrm{x},\:\:\mathrm{y}\right),\:\:\:\:\:\:\mathrm{find}\:\:\:\:\frac{\delta\mathrm{z}}{\delta\mathrm{x}}\:,\:\:\:\frac{\delta\mathrm{z}}{\delta\mathrm{y}} \\ $$ Answered by mr W last updated on 16/Oct/21 $$\mathrm{1}−\frac{\partial{z}}{\partial{x}}=\frac{{y}}{\mathrm{1}+{y}^{\mathrm{2}} {z}^{\mathrm{2}} }×\frac{\partial{z}}{\partial{x}} \\…