Answered

Westonci.ca is the ultimate Q&A platform, offering detailed and reliable answers from a knowledgeable community. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

How do you find the derivative of y=tan(arcsin(x))y=tan(arcsin(x)) ?

Sagot :

Answer is in the attachment below. If you have any questions about the workings, just leave a comment below.
View image Аноним
[tex]y=tan(arcsin(x))=\frac{sin(arcsin(x)}{cos(arcsin(x))}\\\\We\ know:\\sin(arcsin(x))=x\ and\ cos(arcsin(x))=\sqrt{1-x^2}\\\\therefore:y=\frac{x}{\sqrt{1-x^2}}\\\\y'=\left(\frac{x}{\sqrt{1-x^2}}\right)'=\frac{x'\sqrt{1-x^2}-x(\sqrt{1-x^2})'}{(\sqrt{1-x^2})^2}=\frac{\sqrt{1-x^2}-x\cdot\frac{1}{2\sqrt{1-x^2}}\cdot(-2x)}{1-x^2}\\\\=\frac{\sqrt{1-x^2}+\frac{x^2}{\sqrt{1-x^2}}}{1-x^2}=\frac{\frac{1-x^2+x^2}{\sqrt{1-x^2}}}{1-x^2}=\frac{1}{(1-x^2)\sqrt{1-x^2}}=\frac{1}{(1-x^2)(1-x^2)^\frac{1}{2}}[/tex]
[tex].\center\boxed{=\frac{1}{(1-x^2)^\frac{3}{2}}}[/tex]
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.