Find y .
-$$y = - 3 x ^ { 5 } ( 3 x - 4 ) ^ { 2 }$$

Question

Quizplus · Accepted Answer

Using the chain rule, we have:

$$y' = -3(3x-4)^2 \cdot 5x^4 + 5(3x-4)^1 \cdot (-3x^5)$$

Simplifying and factoring out $3x^4$, we have:

$$y' = -15x^4(3x-4)^2$$

Integrating, we obtain:

$$y = \int -15x^4(3x-4)^2 dx$$

Using the substitution $u = 3x-4$, $du = 3dx$, we can rewrite the integral as:

$$y = \frac{-15}{27} \int u^2 \cdot \frac{du}{3} = \frac{-5}{9} \int u^2 du$$

$$y = \frac{-5u^3}{27} + C$$

Undoing the substitution, we get:

$$y = \frac{-5(3x-4)^3}{27} + C$$

Using the initial condition $y(1) = -9720$, we can solve for $C$:

$$-9720 = \frac{-5(3(1)-4)^3}{27} + C$$

$$C = -1134$$

Substituting $C$ back into the equation, we obtain:

$$y = \frac{-5(3x-4)^3}{27} - 1134$$

Expanding this expression, we get:

$$y = \frac{-5(27x^3 - 108x^2 + 144x - 64)}{27} - 1134$$

Simplifying, we obtain:

$$y = \frac{-1134x^3}{3} + 2160x^2 - 960x - 512$$

$$y = - 1134 x ^ { 5 } + 2160 x ^ { 4 } - 960 x ^ { 3 }$$

Therefore, the best choice is B.

Find Y - y=−3x5(3x−4)2y = - 3 x ^ { 5 } ( 3 x - 4 ) ^ { 2 }y=−3x5(3x−4)2