Find the derivative of y=5^(1-4x)
lny=(1-4x)ln5
Can I split it?
lny=ln5-4xln5
1/y dy/dx = -[4ln5/(4xln5)^2]
dy/dx=-[4ln5/(4xln5)^2]y
Thanks.
Can I split it?
lny=ln5-4xln5
Yes, that is valid.
ln y = ln 5 - (4 ln 5) x
since [4 ln 5 is a constant]
(1/y) dy/dx = 0 - (4 ln 5) dx/dx
Thank you!
You're welcome! Yes, you can split the equation.
Starting from the equation lny = ln5 - 4xln5, we want to find the derivative of y with respect to x (dy/dx).
We can use the property of logarithms, which states that ln(a/b) = ln(a) - ln(b), to rewrite ln5 - 4xln5 as ln(5/(5^4x)).
So now the equation becomes lny = ln(5/(5^4x)).
Next, we differentiate both sides of the equation implicitly with respect to x.
On the left side, we use the chain rule:
d/dx(lny) = (1/y) * (dy/dx).
On the right side, we differentiate ln(5/(5^4x)) term by term.
The derivative of ln(5) is 0 because it's a constant with respect to x.
For the derivative of ln(5^4x), we use the chain rule again.
The derivative of ln(u), where u is a function of x, is (1/u) * du/dx. In this case, u = 5^4x.
So, the derivative of ln(5^4x) with respect to x is (1/5^4x) * (d/dx(5^4x)).
Using the chain rule, we have (1/5^4x) * (5^4 * d/dx(x)).
Simplifying further, we get (1/5^4x) * (5^4 * 1).
Therefore, the derivative of ln(5^4x) is 1.
Substituting these values back into our equation, we have:
(1/y) * (dy/dx) = 0 - (1/5^4x).
Multiplying both sides by y, we get:
dy/dx = -(1/5^4x) * y.
Since y = 5^(1-4x), we can substitute its value into the equation:
dy/dx = -(1/5^4x) * 5^(1-4x).
Simplifying further, we have:
dy/dx = -(1/5^4x) * 5^(1) * 5^(-4x).
And finally:
dy/dx = -[4ln5/(4xln5)^2]y.
I hope this explanation helps! Let me know if you have any further questions.