Question

differentiate the function.
$y = (3x - 5)^2 \left(3 - x^4\
ight)^4$
$
\frac{dy}{dx} = \square
$

Explanation:

Step1: Apply product rule

Let $u=(3x-5)^2$ and $v=(3-x^4)^4$. Then $\frac{dy}{dx}=u'v + uv'$.

Step2: Compute $u'$

$u' = 2(3x-5)\cdot3 = 6(3x-5)$ (chain rule).

Step3: Compute $v'$

$v' = 4(3-x^4)^3\cdot(-4x^3) = -16x^3(3-x^4)^3$ (chain rule).

Step4: Substitute back

$\frac{dy}{dx}=6(3x-5)(3-x^4)^4 + (3x-5)^2(-16x^3)(3-x^4)^3$.

Step5: Factor common terms

Factor out $2(3x-5)(3-x^4)^3$:
$\frac{dy}{dx}=2(3x-5)(3-x^4)^3[3(3-x^4) - 8x^3(3x-5)]$.

Step6: Simplify inside brackets

$3(3-x^4) -8x^3(3x-5)=9 - 3x^4 -24x^4 +40x^3=9 +40x^3 -27x^4$.

Answer:

$2(3x - 5)(3 - x^4)^3(9 + 40x^3 - 27x^4)$