Question

differentiate the given function.

$y = x\left(x^4 + 6\
ight)^3$

$dfrac{d}{dx}\left x\left(x^4 + 6\
ight)^3 \
ight = \square$

Explanation:

Step1: Apply Product Rule

The product rule states that if \( y = u \cdot v \), then \( y' = u'v + uv' \). Let \( u = x \) and \( v = (x^4 + 6)^3 \).
First, find \( u' \): \( u = x \), so \( u' = 1 \) (since the derivative of \( x \) with respect to \( x \) is 1).

Step2: Find \( v' \) using Chain Rule

For \( v = (x^4 + 6)^3 \), let \( t = x^4 + 6 \), so \( v = t^3 \).
The chain rule states \( v' = \frac{dv}{dt} \cdot \frac{dt}{dx} \).
First, \( \frac{dv}{dt} = 3t^2 \) (power rule: derivative of \( t^3 \) is \( 3t^2 \)).
Then, \( \frac{dt}{dx} = 4x^3 \) (derivative of \( x^4 \) is \( 4x^3 \), derivative of 6 is 0).
Substitute back \( t = x^4 + 6 \), so \( v' = 3(x^4 + 6)^2 \cdot 4x^3 = 12x^3(x^4 + 6)^2 \).

Step3: Apply Product Rule

Now, \( y' = u'v + uv' = 1 \cdot (x^4 + 6)^3 + x \cdot 12x^3(x^4 + 6)^2 \).
Simplify the second term: \( x \cdot 12x^3 = 12x^4 \), so \( y' = (x^4 + 6)^3 + 12x^4(x^4 + 6)^2 \).
Factor out \( (x^4 + 6)^2 \): \( y' = (x^4 + 6)^2[(x^4 + 6) + 12x^4] \).
Simplify inside the brackets: \( x^4 + 6 + 12x^4 = 13x^4 + 6 \).
So, \( y' = (x^4 + 6)^2(13x^4 + 6) \).

Answer:

\( (x^4 + 6)^2(13x^4 + 6) \)