Find the implicit derivative $\frac{d}{dx}\left(x^4+y^5=\ln\left(\frac{1+y^3}{1-y^3}\right)\right)$

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$\frac{d}{dx}\left(x^4+y^5\right)=\frac{d}{dx}\left(\ln\left(\frac{1+y^3}{1-y^3}\right)\right)$

Learn how to solve problems step by step online. Find the implicit derivative d/dx(x^4+y^5=ln((1+y^3)/(1-y^3))). Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable. The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If f(x)=ln\:a (where a is a function of x), then \displaystyle f'(x)=\frac{a'}{a}. Divide fractions \frac{1}{\frac{1+y^3}{1-y^3}} with Keep, Change, Flip: a\div \frac{b}{c}=\frac{a}{1}\div\frac{b}{c}=\frac{a}{1}\times\frac{c}{b}=\frac{a\cdot c}{b}. Apply the quotient rule for differentiation, which states that if f(x) and g(x) are functions and h(x) is the function defined by {\displaystyle h(x) = \frac{f(x)}{g(x)}}, where {g(x) \neq 0}, then {\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}.