Find the dy/dx for each equation
a) y=ln(2x^6-15)
b) y= 3√2x^2-3x-1/5
c) y=x^4/x^2+1
To find the derivative dy/dx for each equation, we will use the rules of differentiation. Let's take them one by one:
a) y = ln(2x^6 - 15)
To find dy/dx, we will use the chain rule. The chain rule states that if we have a composite function, like ln(u), where u is some expression dependent on x, then the derivative is given by the product of the derivative of the outer function (ln) with the derivative of the inner function.
Let u = 2x^6 - 15
dy/dx = (1/u) * (du/dx)
Now, we need to find du/dx, the derivative of u with respect to x.
du/dx = d/dx (2x^6 - 15) = 12x^5
Plugging this back into the derivative expression for y, we have:
dy/dx = (1/u) * (du/dx)
= (1/(2x^6 - 15)) * (12x^5)
b) y = 3√(2x^2 - 3x - 1)/5
To find dy/dx, we will use the quotient rule. The quotient rule states that if we have a function given as a quotient of two expressions, u(x) and v(x), then the derivative is given by (v(x) * du/dx - u(x) * dv/dx) / (v(x))^2.
Let u(x) = 3√(2x^2 - 3x - 1) and v(x) = 5.
To simplify, let's rewrite u(x) as (2x^2 - 3x - 1)^(1/3).
Now, we can find du/dx and dv/dx.
du/dx = d/dx (2x^2 - 3x - 1)^(1/3)
= (1/3)(2x^2 - 3x - 1)^(-2/3) * d/dx (2x^2 - 3x - 1)
dv/dx = d/dx (5)
= 0 (since the derivative of a constant is zero)
Plugging these values into the quotient rule, we have:
dy/dx = (v(x) * du/dx - u(x) * dv/dx) / (v(x))^2
= [5 * (1/3)(2x^2 - 3x - 1)^(-2/3) * d/dx (2x^2 - 3x - 1)] / 25
c) y = x^4 / (x^2 + 1)
To find dy/dx, we will use the quotient rule again.
Let u(x) = x^4 and v(x) = x^2 + 1.
We need to find du/dx and dv/dx.
du/dx = d/dx (x^4) = 4x^3
dv/dx = d/dx (x^2 + 1) = 2x
Plugging these values into the quotient rule, we have:
dy/dx = (v(x) * du/dx - u(x) * dv/dx) / (v(x))^2
= [(x^2 + 1) * 4x^3 - x^4 * 2x] / (x^2 + 1)^2
And that's how you find the derivatives dy/dx for each equation using the respective differentiation rules.