To solve for $y'$ (or $\frac{dy}{dx}$) in the given implicit equation:

$16x e^y - 18y x^9 + 6y^9 = 5$

we need to differentiate both sides of the equation with respect to $x$, applying implicit differentiation.

Step-by-Step Solution:

1. Differentiate the First Term: $16x e^y$

   • Use the product rule: $\frac{d}{dx}(16x e^y) = 16 e^y + 16x e^y y'$.

2. Differentiate the Second Term: $-18y x^9$

   • Again, apply the product rule: $\frac{d}{dx}(-18y x^9) = -18 x^9 y' - 18y \cdot 9x^8 = -18 x^9 y' - 162y x^8$.

3. Differentiate the Third Term: $6y^9$

   • Use the chain rule: $\frac{d}{dx}(6y^9) = 6 \cdot 9y^8 y' = 54y^8 y'$.

4. Differentiate the Right Side: $5$

   • Since 5 is a constant, its derivative is $0$.

Putting It All Together

After differentiating each term, we get:

$16 e^y + 16x e^y y' - 18 x^9 y' - 162y x^8 + 54y^8 y' = 0$

Now, group all terms involving $y'$ on one side of the equation:

$(16x e^y - 18 x^9 + 54y^8) y' = -16 e^y + 162y x^8$

Then, solve for $y'$:

$y' = \frac{-16 e^y + 162y x^8}{16x e^y - 18 x^9 + 54y^8}$

Answer:

The correct answer is:

$y' = \frac{-16 e^y + 162y x^8}{16x e^y - 18 x^9 + 54y^8}$

Looking at the choices provided in your image, the answer corresponds to the third option.

Would you like further details on any specific step?

Related Questions:

1. How do you apply the product rule in differentiation?
2. What is implicit differentiation, and why is it used here?
3. How does the chain rule apply when differentiating terms involving $y$?
4. What are some typical applications of implicit differentiation in calculus?
5. How would the solution change if the equation involved higher powers of $y$?

Tip:

When differentiating implicitly, keep track of terms involving $y'$ to isolate them easily at the end.