Find the derivative of y=3x-2(4^x)
Here is my attempt:
Derivative of 3x is 3
Derivative of 2 is 0
derivative of 4^x is 4^x(ln(4))
Answer: 12^x(ln(4))
Is this correct? If not please explain what I did wrong and what the correct answer is and why. Thanks!
y=3x-8^x
3-8^x*ln(8)
y = 3x - 2(4^x)
Yep, the derivative of 3x is 3.
Yep, it's true that derivative of 2 is 0, but this is not necessary. Note that 2 is multiplied by 4^x, and does not act as a separate term.
Yep, the derivative of 4^x is 4^x(ln(4)).
Therefore,
y' = 3 - 2*(4^x (ln(4)))
or note that ln(4) = ln(2*2) = ln (2^2) = 2*ln(2). Rewriting,
y' = 3 - 2*2*(4^x (ln(2)))
y' = 3 - 4*(4^x (ln(2)))
y' = 3 - 4^(x+1)*(ln(2))
Hope this helps :)
Your attempt to find the derivative of y = 3x - 2(4^x) is not entirely correct. Let's break down the steps to find the derivative correctly.
First, let's differentiate each term separately:
1. The derivative of 3x is simply 3 since the derivative of x is 1.
2. The derivative of -2(4^x) requires the chain rule. The chain rule states that if we have a composite function, such as f(g(x)), then the derivative of f(g(x)) is f'(g(x)) * g'(x). In this case, -2 is a constant, so its derivative is 0. The derivative of 4^x is (4^x)(ln(4)), which comes from the exponential rule that states the derivative of a^x is (a^x)ln(a). Thus, the derivative of -2(4^x) is -2(4^x)(ln(4)).
Now, let's find the derivative of y = 3x - 2(4^x) by summing up the derivatives of each term:
dy/dx = derivative of (3x) - derivative of (-2(4^x))
= 3 - (-2(4^x)(ln(4)))
Simplifying this expression, we have:
dy/dx = 3 + 2(4^x)(ln(4))
Therefore, the correct answer for the derivative of y = 3x - 2(4^x) is dy/dx = 3 + 2(4^x)(ln(4)).