I need help finding the implicit differentiation of $x^4 +\sin y = x^3y^2$
I especially need help finding $\dfrac{\mathrm d }{\mathrm dx}$for both sides. I'm having trouble with the product rule on the right side with $\dfrac{\mathrm d y}{\mathrm dx}$ of $x^3 y^2$.
[Math] Implicit differentiation of $x^4 +\sin y = x^3y^2$
calculus
Best Answer
I think it helps if you think of $y$ as a function of $x$, and in fact write $y=f(x)$. Then the product rule says $$\frac{d}{dx}(x^3y^2)=\frac{d}{dx}(x^3 f(x)^2)=\frac{d}{dx}(x^3)\cdot f(x)^2 + x^3\cdot \frac{d}{dx}(f(x)^2)$$ In order to continue, you need to calculate $\frac{d}{dx}(f(x)^2)$ using the chain rule, and the result is $2\cdot f(x)\cdot \frac{d}{dx}(f(x))$. Hence, $$\frac{d}{dx}(x^3y^2)=\frac{d}{dx}(x^3)\cdot f(x)^2 + x^3\cdot \frac{d}{dx}(f(x)^2)=3x^2f(x)^2+x^3\cdot 2\cdot f(x)\cdot \frac{d}{dx}(f(x)).$$ Going back to the notation $y=f(x)$, and $\frac{dy}{dx}=\frac{d}{dx}(f(x))$, we obtain $$\frac{d}{dx}(x^3y^2)=3x^2y^2+2x^3y\frac{dy}{dx}.$$