## Calculus 10th Edition

$\dfrac{dy}{dx}=-\dfrac{y^2}{{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}}.$
$\dfrac{d}{dx}(x)=\dfrac{d}{dx}(\sec{\dfrac{1}{y}})\rightarrow$ Using the Chain Rule with $u=y^{-1}\rightarrow\dfrac{du}{dx}=-\dfrac{1}{y^2}\rightarrow$ $\dfrac{d}{dx}(\sec{\dfrac{1}{y}})=\dfrac{dy}{dx}(-\dfrac{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}{y^2})\rightarrow$ $1=\dfrac{dy}{dx}(-\dfrac{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}{y^2})\rightarrow$ $\dfrac{dy}{dx}=-\dfrac{y^2}{{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}}.$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}