find dy/dx by implicit differentiation for the following equation. x^5y^3 + 8x^(5/4) = 6y^(5/2) + 5 dy/dx = □

Explanation:

Step1: Differentiate each term

Differentiate $x^{5}y^{3}$ using product - rule $(uv)^\prime = u^\prime v+uv^\prime$ where $u = x^{5}$ and $v = y^{3}$. The derivative of $x^{5}$ with respect to $x$ is $5x^{4}$, and the derivative of $y^{3}$ with respect to $x$ is $3y^{2}\frac{dy}{dx}$. So, $\frac{d}{dx}(x^{5}y^{3})=5x^{4}y^{3}+3x^{5}y^{2}\frac{dy}{dx}$. The derivative of $8x^{\frac{5}{4}}$ with respect to $x$ is $8\times\frac{5}{4}x^{\frac{5}{4}-1}=10x^{\frac{1}{4}}$. The derivative of $6y^{\frac{5}{2}}$ with respect to $x$ is $6\times\frac{5}{2}y^{\frac{5}{2}-1}\frac{dy}{dx}=15y^{\frac{3}{2}}\frac{dy}{dx}$, and the derivative of the constant 5 with respect to $x$ is 0. So, $\frac{d}{dx}(x^{5}y^{3}+8x^{\frac{5}{4}})=\frac{d}{dx}(6y^{\frac{5}{2}} + 5)$ gives $5x^{4}y^{3}+3x^{5}y^{2}\frac{dy}{dx}+10x^{\frac{1}{4}}=15y^{\frac{3}{2}}\frac{dy}{dx}+0$.

Step2: Isolate $\frac{dy}{dx}$

Move all terms with $\frac{dy}{dx}$ to one side: $3x^{5}y^{2}\frac{dy}{dx}-15y^{\frac{3}{2}}\frac{dy}{dx}=- 5x^{4}y^{3}-10x^{\frac{1}{4}}$. Factor out $\frac{dy}{dx}$: $\frac{dy}{dx}(3x^{5}y^{2}-15y^{\frac{3}{2}})=-5x^{4}y^{3}-10x^{\frac{1}{4}}$. Then $\frac{dy}{dx}=\frac{-5x^{4}y^{3}-10x^{\frac{1}{4}}}{3x^{5}y^{2}-15y^{\frac{3}{2}}}=\frac{5x^{4}y^{3} + 10x^{\frac{1}{4}}}{15y^{\frac{3}{2}}-3x^{5}y^{2}}$.

Answer:

$\frac{5x^{4}y^{3}+10x^{\frac{1}{4}}}{15y^{\frac{3}{2}} - 3x^{5}y^{2}}$