Differentiate: y = 3/(1-6x^4)
I would write it as
y = 3(1 - 6x^4)^-1 , and use the chain rule
dy/dx = -3(1 - 6x^4)^-2 (-24x^3
= 72x^3/(1 - 6x^4)^2
To differentiate the function y = 3/(1 - 6x^4), you can use the quotient rule. The quotient rule states that if you have a function of the form f(x) = g(x)/h(x), where g(x) and h(x) are differentiable functions, then the derivative of f(x) is given by:
f'(x) = (h(x) * g'(x) - g(x) * h'(x)) / (h(x))^2
Now, let's find the derivative of the given function step-by-step:
Step 1: Identify g(x) and h(x) in the form f(x) = g(x)/h(x).
g(x) = 3
h(x) = 1 - 6x^4
Step 2: Find g'(x) and h'(x).
g'(x) = 0 (since the derivative of a constant is zero)
h'(x) = -24x^3 (using the power rule for differentiation)
Step 3: Plug g(x), h(x), g'(x), and h'(x) into the quotient rule formula:
f'(x) = (h(x) * g'(x) - g(x) * h'(x)) / (h(x))^2
= ((1 - 6x^4) * 0 - 3 * (-24x^3)) / (1 - 6x^4)^2
= (0 + 72x^3) / (1 - 6x^4)^2
= 72x^3 / (1 - 6x^4)^2
So, the derivative of y = 3/(1 - 6x^4) is y' = 72x^3 / (1 - 6x^4)^2.