Given the equation [tex]\( y^3 = 27x^3 + 54x^2 + 36x + 8 \)[/tex], we want to find the derivative [tex]\(\frac{dy}{dx}\)[/tex].
First, let's rewrite the given equation:
[tex]\[ y^3 = 27x^3 + 54x^2 + 36x + 8 \][/tex]
We need to implicitly differentiate both sides of this equation with respect to [tex]\(x\)[/tex].
Differentiating the left side with respect to [tex]\(x\)[/tex] using the chain rule:
[tex]\[ \frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx} \][/tex]
Next, differentiate the right side with respect to [tex]\(x\)[/tex]:
[tex]\[ \frac{d}{dx}(27x^3 + 54x^2 + 36x + 8) = 81x^2 + 108x + 36 \][/tex]
Equating both derivatives, we have:
[tex]\[ 3y^2 \frac{dy}{dx} = 81x^2 + 108x + 36 \][/tex]
Now, solving for [tex]\(\frac{dy}{dx}\)[/tex]:
[tex]\[ \frac{dy}{dx} = \frac{81x^2 + 108x + 36}{3y^2} \][/tex]
Since [tex]\(y = (27x^3 + 54x^2 + 36x + 8)^{1/3}\)[/tex], [tex]\(y^2 = \left( (27x^3 + 54x^2 + 36x + 8)^{1/3} \right)^2 = (27x^3 + 54x^2 + 36x + 8)^{2/3}\)[/tex].
Substituting [tex]\(y^2\)[/tex] back into the equation:
[tex]\[ \frac{dy}{dx} = \frac{81x^2 + 108x + 36}{3(27x^3 + 54x^2 + 36x + 8)^{2/3}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{dy}{dx} = \frac{81x^2 + 108x + 36}{(27x^3 + 54x^2 + 36x + 8)^{2/3}} \times \frac{1}{3} \][/tex]
So, factoring out the common factor in the numerator:
[tex]\[ \frac{dy}{dx} = \frac{27(x^2 + 1.333\ldots x + 0.444\ldots)}{(27x^3 + 54x^2 + 36x + 8)^{2/3}} \times \frac{1}{3} \][/tex]
Combining the constants:
[tex]\[ \frac{dy}{dx} = \frac{27(x^2 + \frac{4}{3}x + \frac{4}{9})}{(27x^3 + 54x^2 + 36x + 8)^{2/3}} \times \frac{1}{3} \][/tex]
Finally, simplifying, we get:
[tex]\[ \frac{dy}{dx} = \frac{27x^2 + 36x + 12}{(27x^3 + 54x^2 + 36x + 8)^{2/3}} \][/tex]
Therefore:
[tex]\[ \frac{dy}{dx} = \frac{27.0x^2 + 36.0x + 12.0}{(27x^3 + 54x^2 + 36x + 8)^{0.666666666666667}} \][/tex]
