y=3x-8^x
3-8^x*ln(8)
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!
2 answers
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 :)
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 :)