Use logarithmic differentiation to find the derivative of the following equation.
y = (2x+1)^5(x^4−3)^6
y=(2x+1)^5*(x^4-3)^6 {take the ln of both sides}
ln(y) = ln((2x+1)^5*(x^4-3)^6) {simplify the right side}
ln(y) = ln((2x+1)^5) + ln((x^4-3)^6) {simplify further}
ln(y) = 5ln(2x+1) + 6ln(x^4-3) {take the derivative} {lets do each term individually}
Deriviative of ln(y) =(1/y)y' {deriviative of ln(y) = 1/y}
Deriviative of 5ln(2x+1) = 5*2/(2x+1) = 10/(2x+1) {Use chain rule, let u=2x+1; du=2}
Deriviative of 6ln(x^4-3) = 6*4x^3/(x^4-3) = 24x^3/(x^4-3) {Use chain rule, let u=x^4-3; du=4x^3}
{now put it together}
(1/y)y' = 10/(2x+1) + 24x^3/(x^4-3) {multiply both sides by y}
y' = y * (10/(2x+1) + 24x^3/(x^4-3)) {sustitute y for (2x+1)^5*(x^4-3)^6 (the original equation)}
y' = (2x+1)^5*(x^4-3)^6 * (10/(2x+1) + 24x^3/(x^4-3)) {simplify}
y' = 10(2x+1)^4*(x^4-3)^6 + (24x^3)(2x+1)^5*(x^4-3)^5 {that's about as far as you go}
you can factor out the 2(2x+1)^4 (x^4-3)^5
and you have
2(2x+1)^4 (x^4-3)^5 (5(x^4-3) + 12x^3(2x+1))
= 2(2x+1)^4 (x^4-3)^5 (29x^4+12x^3-15)
Well, logarithmic differentiation sounds like a complicated method, but don't worry, I'm here to make it less intimidating with a touch of humor!
Let's start by taking the natural logarithm of both sides:
ln(y) = ln[(2x+1)^5(x^4−3)^6]
Now, let me clarify one thing: taking derivatives of logarithms is easier than pronouncing the word "logarithmic differentiation" five times fast. Trust me on this!
Using the logarithmic property, we can break down the expression into two parts:
ln(y) = 5ln(2x+1) + 6ln(x^4−3)
Now, it's time for the fun part! Differentiate both sides using the chain rule, like a circus acrobat:
(dy/dx) / y = 5(2/(2x+1)) + 6(4x^3/(x^4-3))
See, it wasn't as scary as eating a clown's cooking! Now, let's simplify a bit more:
(dy/dx) / y = 10/(2x+1) + 24x^3/(x^4-3)
Now, we shall get rid of that pesky denominator by multiplying both sides by y:
dy/dx = y(10/(2x+1) + 24x^3/(x^4-3))
But don't forget, we want to express the derivative in terms of x, not y. So let's substitute y back in:
dy/dx = (2x+1)^5(x^4−3)^6 [10/(2x+1) + 24x^3/(x^4-3)]
And there you have it! The derivative of the original equation using logarithmic differentiation. Now, go spread some laughter in your calculus class!
To find the derivative of the equation y = (2x+1)^5(x^4−3)^6 using logarithmic differentiation, follow these steps:
Step 1: Take the natural logarithm of both sides of the equation:
ln(y) = ln((2x+1)^5(x^4−3)^6)
Step 2: Apply the logarithm rules to simplify the equation:
ln(y) = 5ln(2x+1) + 6ln(x^4−3)
Step 3: Differentiate both sides of the equation with respect to x using the chain rule:
(1/y) * dy/dx = 5(1/(2x+1)) * (d(2x+1)/dx) + 6(1/(x^4−3)) * (d(x^4−3)/dx)
Step 4: Simplify the derivatives:
(1/y) * dy/dx = 5(1/(2x+1)) * 2 + 6(1/(x^4−3)) * (4x^3)
Step 5: Simplify further:
(1/y) * dy/dx = 10/(2x+1) + 24x^3/(x^4−3)
Step 6: Multiply both sides of the equation by y to solve for dy/dx:
dy/dx = y * (10/(2x+1) + 24x^3/(x^4−3))
Step 7: Substitute back the original expression for y:
dy/dx = (2x+1)^5(x^4−3)^6 * (10/(2x+1) + 24x^3/(x^4−3))
Therefore, the derivative of the equation y = (2x+1)^5(x^4−3)^6 using logarithmic differentiation is dy/dx = (2x+1)^5(x^4−3)^6 * (10/(2x+1) + 24x^3/(x^4−3)).
To find the derivative of the given equation using logarithmic differentiation, follow these steps:
Step 1: Take the natural logarithm of both sides of the equation.
ln(y) = ln((2x+1)^5(x^4−3)^6)
Step 2: Apply the logarithmic rules to simplify the equation.
Using the properties of logarithms, we can break down the equation as follows:
ln(y) = 5ln(2x+1) + 6ln(x^4−3)
Step 3: Differentiate both sides of the equation with respect to x.
Using the chain rule, we can differentiate each term on the right-hand side separately.
d/dx(ln(y)) = d/dx(5ln(2x+1)) + d/dx(6ln(x^4−3))
Step 4: Simplify the derivatives on the right-hand side of the equation.
To differentiate the first term, 5ln(2x+1), we can use the chain rule. Let's denote it as u.
Let u = 2x+1
Then, ln(u) = ln(2x+1)
Now, we can differentiate both sides with respect to x using the chain rule.
d/dx(ln(u)) = d/dx(ln(2x+1))
1/u(du/dx) = 1/(2x+1)(d(2x+1)/dx)
1/u * 1 = 1/(2x+1) * 2
Simplifying this gives:
du/dx = 2/(2x+1)
Thus, the derivative of 5ln(2x+1) is:
d(5ln(2x+1))/dx = 5 * (2/(2x+1)) = 10/(2x+1)
Next, we differentiate the second term, 6ln(x^4−3), using similar steps.
Let v = x^4−3
Then, ln(v) = ln(x^4−3)
Differentiating both sides with respect to x using the chain rule:
d/dx(ln(v)) = d/dx(ln(x^4−3))
1/v(dv/dx) = 1/(x^4-3)(d(x^4-3)/dx)
Expanding and simplifying:
1/v * 1 = 1/(x^4-3) * 4x^3
dv/dx = 4x^3/(x^4-3)
Hence, the derivative of 6ln(x^4-3) is:
d(6ln(x^4-3))/dx = 6 * (4x^3/(x^4-3)) = 24x^3/(x^4-3)
Step 5: Combine the derivatives obtained from Step 4.
d/dx(ln(y)) = 10/(2x+1) + 24x^3/(x^4-3)
Step 6: Now, we need to find dy/dx, the derivative of y with respect to x.
To do this, we multiply both sides of the equation obtained in Step 6 by y.
dy/dx * (1/y) = (10/(2x+1) + 24x^3/(x^4-3)) * y
The left-hand side simplifies to:
dy/dx * (1/y) = dy/dx * (1/y) = dy/dx * (y^-1) = d/dx(y^-1) = -y^-2 * dy/dx
Hence, we have:
-y^-2 * dy/dx = (10/(2x+1) + 24x^3/(x^4-3)) * y
Step 7: Solve for dy/dx.
Multiply both sides of the equation by -y^2 to isolate dy/dx:
dy/dx = -y^2 * (10/(2x+1) + 24x^3/(x^4-3)) * y
Finally, substitute the original equation for y:
dy/dx = -y^2 * (10/(2x+1) + 24x^3/(x^4-3)) * y = -(2x+1)^5(x^4−3)^6 * (10/(2x+1) + 24x^3/(x^4-3))
Therefore, the derivative of the equation y = (2x+1)^5(x^4−3)^6 using logarithmic differentiation is:
dy/dx = -(2x+1)^5(x^4−3)^6 * (10/(2x+1) + 24x^3/(x^4-3))