Question

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differentiate.
( y = \frac{5}{1 - 3x^4} )

Explanation:

Step1: Rewrite the function

Rewrite \( y = \frac{5}{1 - 3x^4} \) as \( y = 5(1 - 3x^4)^{-1} \) (using the negative exponent rule \( \frac{1}{a^n}=a^{-n} \)).

Step2: Apply the chain rule

The chain rule states that if \( y = f(g(x)) \), then \( y' = f'(g(x))\cdot g'(x) \). Let \( u = 1 - 3x^4 \), so \( y = 5u^{-1} \). First, differentiate \( y \) with respect to \( u \): \( \frac{dy}{du}=5\times(-1)u^{-2}=-5u^{-2} \). Then differentiate \( u \) with respect to \( x \): \( \frac{du}{dx}=-12x^3 \) (using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), here \( n = 4 \), so derivative of \( - 3x^4 \) is \( - 3\times4x^{3}=-12x^{3} \), derivative of 1 is 0).

Step3: Multiply the derivatives

By the chain rule, \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \). Substitute \( u = 1 - 3x^4 \), \( \frac{dy}{du}=-5(1 - 3x^4)^{-2} \) and \( \frac{du}{dx}=-12x^3 \) into it: \( \frac{dy}{dx}=-5(1 - 3x^4)^{-2}\times(-12x^3)=60x^3(1 - 3x^4)^{-2}=\frac{60x^3}{(1 - 3x^4)^{2}} \) (or we can also simplify the given wrong answer's format, but the correct derivative is calculated as above. Wait, the given answer in the box seems to have a mistake in the exponent. Let's recalculate properly.

Wait, maybe the original function was \( y=\frac{5}{(1 - 3x^4)^5} \) (maybe a typo in the problem statement? Because the given derivative has exponent -6). Assuming the function is \( y = \frac{5}{(1 - 3x^4)^5}=5(1 - 3x^4)^{-5} \). Then:

Step1: Rewrite the function

\( y = 5(1 - 3x^4)^{-5} \)

Step2: Apply chain rule

Let \( u = 1 - 3x^4 \), \( y = 5u^{-5} \). \( \frac{dy}{du}=5\times(-5)u^{-6}=-25u^{-6} \), \( \frac{du}{dx}=-12x^3 \)

Step3: Multiply derivatives

\( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=-25(1 - 3x^4)^{-6}\times(-12x^3)=300x^3(1 - 3x^4)^{-6}=\frac{300x^3}{(1 - 3x^4)^6} \). But the given box has \( - 5(1 - 3x^4)^{-6}\cdot(12x^3) \), which is \( - 60x^3(1 - 3x^4)^{-6} \), which is wrong. Maybe the original function was \( y=\frac{5}{(1 - 3x^4)^5} \) and there was a miscalculation in the coefficient. But if we follow the problem as given (maybe a typo in the exponent of the denominator), let's assume the function is \( y=\frac{5}{(1 - 3x^4)^5} \), then the correct derivative steps:

But the user's problem is to differentiate \( y = \frac{5}{1 - 3x^4} \) (as per the written problem). Let's do that correctly:

Step1: Rewrite

\( y = 5(1 - 3x^4)^{-1} \)

Step2: Differentiate using chain rule

\( \frac{dy}{dx}=5\times(-1)(1 - 3x^4)^{-2}\times(-12x^3) \) (derivative of outer function \( 5u^{-1} \) is \( - 5u^{-2} \), derivative of inner function \( u = 1 - 3x^4 \) is \( - 12x^3 \))

Step3: Simplify

\( \frac{dy}{dx}=60x^3(1 - 3x^4)^{-2}=\frac{60x^3}{(1 - 3x^4)^2} \)

But the given answer in the box has a different exponent, so maybe there was a typo in the original function (like denominator is \( (1 - 3x^4)^5 \) instead of \( 1 - 3x^4 \)). Assuming the function is \( y=\frac{5}{(1 - 3x^4)^5} \), then:

Step1: Rewrite

\( y = 5(1 - 3x^4)^{-5} \)

Step2: Chain rule

\( \frac{dy}{dx}=5\times(-5)(1 - 3x^4)^{-6}\times(-12x^3) \)

Step3: Simplify

\( \frac{dy}{dx}=300x^3(1 - 3x^4)^{-6}=\frac{300x^3}{(1 - 3x^4)^6} \)

But the given box has \( - 5(1 - 3x^4)^{-6}\cdot(12x^3) \), which is \( - 60x^3(1 - 3x^4)^{-6} \), which is incorrect. Maybe the original function was \( y = \frac{5}{(1 - 3x^4)^5} \) and the coefficient was miscalculated. However, if we follow the problem as written ( \( y=\frac{5}{1 - 3x^4} \) ), the correct derivative is \( \frac{60x^3}{(1 - 3x^4)^2} \).

Answer:

If the function is \( y = \frac{5}{1 - 3x^4} \), the derivative \( \frac{dy}{dx}=\frac{60x^3}{(1 - 3x^4)^2} \). If there was a typo and the function is \( y=\frac{5}{(1 - 3x^4)^5} \), the derivative is \( \frac{300x^3}{(1 - 3x^4)^6} \) (correcting the coefficient error in the given box).