Animated Solution for Mathematics - Differentiation: Which of the following points lies on the tangent to the curve $x^4{e}^y+2\sqrt{y}+1=3$ at the point $(1,0)$?

Unmasking the Curve

Imagine you are standing on a path defined by the equation $x^4{e}^y+2\sqrt{y+1}=3$. You are asked to find the tangent line at the point $(1,0)$.

Before you even pick up your pen to differentiate, you must act like a detective. Does the point $(1,0)$ actually sit on this path?

If we plug $x=1$ and $y=0$ into the original equation, we find that the math fits perfectly: $1^4{e}^0+2\sqrt{0+1}=1(1)+2(1)=3$. Always verify your coordinates before you start the heavy lifting!

Implicit Differentiation

Now, look at the equation. The variables $x$ and $y$ are locked in a complex embrace, and we cannot easily write $y=f(x)$.

This is where we call upon our most reliable weapon: Implicit Differentiation. We treat $y$ as a function of $x$ and differentiate both sides with respect to $x$.

For the first term, $x^4{e}^y$, we must use the Product Rule: $\frac{d}{dx}(uv)=u^{\prime }v+u{v}^{\prime }$. The derivative of $x^4$ is $4x^3$, and the derivative of $e^y$ is $e^y\cdot \frac{dy}{dx}$ (thanks to the Chain Rule).

So, the derivative of the first term becomes:

Next, we tackle $2\sqrt{y+1}$. The derivative of $\sqrt{u}$ is $\frac{1}{2\sqrt{u}}$.

Applying this, the derivative of $2\sqrt{y+1}$ becomes $2\cdot \frac{1}{2\sqrt{y+1}}\cdot \frac{dy}{dx}$, which simplifies beautifully to:

Now, combine everything. The derivative of the constant $3$ on the right side is simply $0$. Our full differentiated equation is:

Substitute Early

Here is where most students lose time. They try to isolate $\frac{dy}{dx}$ algebraically while the equation is still a mess. Don't do that!

We only need the slope at the specific point $(1,0)$. So, substitute $x=1$ and $y=0$ right now. The equation collapses:

Since $e^0=1$ and $\sqrt{1}=1$, this becomes $4+(1+1)\frac{dy}{dx}=0$, or simply $4+2\frac{dy}{dx}=0$. Solving for the slope $m$, we get:

We have the point $(1,0)$ and the slope $m=-2$. Using the point-slope form $y-y_1=m(x-x_1)$, we get $y-0=-2(x-1)$.

This simplifies to $y=-2x+2$, or $2x+y=2$.

Now, check the options. For the point $(-2,6)$, we calculate $2(-2)+6=-4+6=2$. It matches perfectly!

You have successfully navigated the trap, mastered the implicit differentiation, and arrived at the correct answer. Keep this mindset—verify, differentiate, substitute, and solve—and no JEE problem will ever intimidate you again.