Animated Solution for Mathematics - Differentiation: If $x{e}^{xy}=y+{\sin }^{2}x$, then at $x=0,\frac{dy}{dx}=\dots \dots \dots$

Imagine you are standing on a complex, winding path defined by the equation $x{e}^{xy}=y+{\sin }^{2}x$. You want to know the slope of the path exactly at the point where $x=0$.

When we face an equation where $y$ is not easily isolated, we use the power of implicit differentiation. Let us break this down step by step.

Before we can find the slope, we need to know exactly where we are on this curve. We know $x=0$, so we substitute this into our original equation:

Since $0$ times anything is $0$, and $\sin (0)=0$, the equation simplifies beautifully to $0=y+0$, which means $y=0$.

Our point of evaluation is the origin, $(0,0)$. We are looking for the slope of the tangent line at the very center of the coordinate system.

Now, we differentiate both sides with respect to $x$. On the left, we have $x\cdot e^{xy}$. This is a product of two functions, so we must use the product rule: $(uv{)}^{\prime }=u^{\prime }v+u{v}^{\prime }$.

Here, $u=x$ and $v=e^{xy}$. The derivative of $x$ is $1$. The derivative of $e^{xy}$ requires the chain rule: it is $e^{xy}$ multiplied by the derivative of the exponent $xy$.

Using the product rule again on $xy$, the derivative is $y+x\frac{dy}{dx}$. Putting it all together, the derivative of the left side is:

The right side, $y+{\sin }^{2}x$, is much friendlier. The derivative of $y$ is simply $\frac{dy}{dx}$.

For ${\sin }^{2}x$, we use the chain rule: the derivative of $(\sin x{)}^2$ is $2\sin x\cdot \cos x$. So, the derivative of the right side is:

Now, we equate the two sides:

Many students would now try to rearrange this massive equation to solve for $\frac{dy}{dx}$. But look at the structure! We only need the value at $(0,0)$.

Let us substitute $x=0$ and $y=0$ immediately. The term $e^{xy}$ becomes $e^0=1$. The term $x{e}^{xy}(y+x\frac{dy}{dx})$ becomes $0\cdot 1\cdot (0+0\cdot \frac{dy}{dx})$, which is $0$.

On the right side, $2\sin (0)\cos (0)$ is also $0$. The entire equation collapses into:

Thus, $\frac{dy}{dx}=1$. The slope of the tangent line at the origin is exactly $1$. By choosing to substitute early, we turned a terrifying algebraic nightmare into a moment of pure, elegant simplicity.