Answer:
[tex]\dfrac{\text{d}y}{\text{d}x}=-\dfrac{15y}{5x+6\sqrt[3]{y^2}}[/tex]
Step-by-step explanation:
Given equation:
[tex]5x \sqrt[3]{y} + 2y=26[/tex]
To differentiate an equation that contains a mixture of x and y terms, we can use implicit differentiation.
Begin by placing d/dx in front of each term of the given equation:
[tex]\dfrac{\text{d}}{\text{d}x}\:5x\sqrt[3]{y}+\dfrac{\text{d}}{\text{d}x}\:2y=\dfrac{\text{d}}{\text{d}x}\:26[/tex]
Differentiate the constant term:
[tex]\dfrac{\text{d}}{\text{d}x}\:5x\sqrt[3]{y}+\dfrac{\text{d}}{\text{d}x}\:2y=0[/tex]
Use the chain rule to differentiate the term in y only.
In practice, this means differentiate with respect to y, and place dy/dx at the end:
[tex]\dfrac{\text{d}}{\text{d}x}\:5x\sqrt[3]{y}+2\:\dfrac{\text{d}y}{\text{d}x}=0[/tex]
Use the product rule to differentiate the term in x and y.
[tex]\boxed{\begin{array}{c}\underline{\textsf{Product Rule for Differentiation}}\\\\\textsf{If}\;y=uv\;\textsf{then:}\\\\\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}\end{array}}[/tex]
[tex]\textsf{Let $u$}=5x \implies \dfrac{\text{d}u}{\text{d}x}=5[/tex]
[tex]\textsf{Let $v$}=\sqrt[3]{y}=y^{\frac{1}{3}} \implies \dfrac{\text{d}v}{\text{d}x}=\dfrac{1}{3}y^{-\frac23}\cdot\dfrac{\text{d}y}{\text{d}x}[/tex]
(Again, use the chain rule to differentiate v with respect to x, which means differentiate with respect to y and place dy/dx at the end).
Substitute everything into the product rule formula:
[tex]\dfrac{\text{d}}{\text{d}x}\:5x\sqrt[3]{y}=5x \cdot \dfrac{1}{3}y^{-\frac23}\cdot\dfrac{\text{d}y}{\text{d}x}+\sqrt[3]{y} \cdot 5 \\\\\\ \dfrac{\text{d}}{\text{d}x}\:5x\sqrt[3]{y}=\dfrac{5x}{3\sqrt[3]{y^2}\;}\dfrac{\text{d}y}{\text{d}x}+5\sqrt[3]{y}[/tex]
Now substitute this into the original differentiated equation:
[tex]\dfrac{5x}{3\sqrt[3]{y^2}\;}\dfrac{\text{d}y}{\text{d}x}+5\sqrt[3]{y}+2\:\dfrac{\text{d}y}{\text{d}x}=0[/tex]
Rearrange to isolate dy/dx:
[tex]\dfrac{5x}{3\sqrt[3]{y^2}}\:\dfrac{\text{d}y}{\text{d}x}+2\:\dfrac{\text{d}y}{\text{d}x}=-5\sqrt[3]{y}\\\\\\ \dfrac{\text{d}y}{\text{d}x}\left(\dfrac{5x}{3\sqrt[3]{y^2}}+2\right)=-5\sqrt[3]{y} \\\\\\ \dfrac{\text{d}y}{\text{d}x}\left(\dfrac{5x}{3\sqrt[3]{y^2}}+\dfrac{6\sqrt[3]{y^2}}{3\sqrt[3]{y^2}}\right)=-5\sqrt[3]{y}\\\\\\ \dfrac{\text{d}y}{\text{d}x}\left(\dfrac{5x+6\sqrt[3]{y^2}}{3\sqrt[3]{y^2}}\right)=-5\sqrt[3]{y}[/tex]
[tex]\dfrac{\text{d}y}{\text{d}x}\left(5x+6\sqrt[3]{y^2}}\right)=-5\sqrt[3]{y} \cdot 3\sqrt[3]{y^2}\\\\\\ \dfrac{\text{d}y}{\text{d}x}\left(5x+6\sqrt[3]{y^2}}\right)=-15\sqrt[3]{y\cdot y^2} \\\\\\ \dfrac{\text{d}y}{\text{d}x}\left(5x+6\sqrt[3]{y^2}}\right)=-15\sqrt[3]{y^3}\\\\\\ \dfrac{\text{d}y}{\text{d}x}\left(5x+6\sqrt[3]{y^2}}\right)=-15y\\\\\\ \dfrac{\text{d}y}{\text{d}x}=-\dfrac{15y}{5x+6\sqrt[3]{y^2}}[/tex]
Therefore, the derivative of the given equation is:
[tex]\Large\boxed{\dfrac{\text{d}y}{\text{d}x}=-\dfrac{15y}{5x+6\sqrt[3]{y^2}}}[/tex]