For the equation \( x^{\frac{3}{5}}+y^{\frac{1}{2}}=7 \). Find the equation of the tangent line at the point \( (1,36) \). (Use symbolic notation and fractions where needed.) Equation of the tangent line is \( y=\square \) help (fractions)

Given the equation \[ x^{\frac{3}{5}} + y^{\frac{1}{2}} = 7, \] we wish to find the equation of the tangent line at the point \((1, 36)\). 1. Differentiate both sides with respect to \(x\). Using the chain rule for the second term, we have \[ \frac{d}{dx}\left(x^{\frac{3}{5}}\right) + \frac{d}{dx}\left(y^{\frac{1}{2}}\right) = 0. \] The derivative of the first term is \[ \frac{3}{5} x^{-\frac{2}{5}}, \] and for the second term (applying the chain rule), \[ \frac{1}{2} y^{-\frac{1}{2}} \cdot y' . \] Thus, the differentiated equation becomes \[ \frac{3}{5} x^{-\frac{2}{5}} + \frac{1}{2} y^{-\frac{1}{2}}\, y' = 0. \] 2. Solve for \(y'\): \[ \frac{1}{2} y^{-\frac{1}{2}}\, y' = -\frac{3}{5} x^{-\frac{2}{5}}, \] hence, \[ y' = -\frac{3}{5} x^{-\frac{2}{5}} \cdot 2 y^{\frac{1}{2}} = -\frac{6}{5} x^{-\frac{2}{5}} y^{\frac{1}{2}}. \] 3. Evaluate \(y'\) at the point \((1,36)\). Notice that \[ 1^{-\frac{2}{5}} = 1 \quad \text{and} \quad 36^{\frac{1}{2}} = 6. \] Substituting gives: \[ y' = -\frac{6}{5} \cdot 1 \cdot 6 = -\frac{36}{5}. \] This is the slope \(m\) of the tangent line. 4. Write the equation of the tangent line using the point-slope form: \[ y - 36 = -\frac{36}{5}(x - 1). \] Simplifying, we have: \[ y = -\frac{36}{5} x + \frac{36}{5} + 36. \] Express \(36\) as \(\frac{180}{5}\) to combine the constants: \[ y = -\frac{36}{5} x + \frac{36 + 180}{5} = -\frac{36}{5} x + \frac{216}{5}. \] Thus, the equation of the tangent line is \[ y = -\frac{36}{5}x + \frac{216}{5}. \]