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Question Number 72343 by Rio Michael last updated on 27/Oct/19

$$giventhaty=ln(1+{cos}^{2}x)find\frac{dy}{dx}atthepointx=\frac{3\pi }{4}$$$$andify=ln(x^2+4)find\frac{dy}{dx}atx=1$$

Commented by mathmax by abdo last updated on 27/Oct/19

$$y\left(x\right)=ln(1+{cos}^{2}x)\Rightarrow y^{{}^{\prime }}\left(x\right)=\frac{-2sinxcosx}{1+{cos}^{2}x}\Rightarrow$$$$y^{{}^{\prime }}\left(\frac{3\pi }{4}\right)=\frac{-2sin\left(\frac{3\pi }{4}\right)cos\left(\frac{3\pi }{4}\right)}{1+{cos}^2\left(\frac{3\pi }{4}\right)}=\frac{2sin\left(\frac{\pi }{4}\right)cos\left(\frac{\pi }{4}\right)}{1+{cos}^2\left(\frac{\pi }{4}\right)}=\frac{2.\frac{1}{\sqrt{2}}.\frac{1}{\sqrt{2}}}{1+{\left(\frac{1}{\sqrt{2}}\right)}^2}$$$$=\frac{1}{\frac{3}{2}}=\frac{2}{3}$$$$y\left(x\right)=ln(x^2+4)\Rightarrow y^{{}^{\prime }}\left(x\right)=\frac{2x}{x^2+4}\Rightarrow y^{{}^{\prime }}\left(1\right)=\frac{2}{5}$$

Commented by Rio Michael last updated on 30/Oct/19

$$thankyousir$$

Commented by mathmax by abdo last updated on 31/Oct/19

$$youarewelcome.$$

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