How Do You Differentiate the Function y = 2^x + x - 4?

[mr bob]

Just came across a question today with 2^x and realized i didnt know how to differentiate it. The entire function i had to differentiate was
[math]y = 2^x + x -4[math]

I tried taking logs but couldn't get anywhere near the true answer.

What is the correct method for this?

Thank's alot!

 

[arildno]

Rewrite 2^{x}=e^{xlog(2)} and use the chain rule on your right-hand side in order to find an expression for \frac{d}{dx}2^{x}

 

[Jeff Ford]

Do you know the rule for the derivative of an expression a^x?

 

[physicscrap]

Just came across a question today with 2^x and realized i didnt know how to differentiate it. The entire function i had to differentiate was
[math]y = 2^x + x -4[math]

I tried taking logs but couldn't get anywhere near the true answer.

What is the correct method for this?

Thank's alot!

y = a^u

y' = lna * a^u * u'

y = 2^x y' = ln2 * 2^x * 1
derivative of (x-4) is 1

so the answer is y' = 2^x * ln2 + 1

 

[VietDao29]

y = a^u

y' = lna * a^u * u'

y = 2^x y' = ln2 * 2^x * 1
derivative of (x-4) is 1

so the answer is y' = 2^x * ln2 + 1

DO NOT EVER POST COMPLETE SOLUTION. THE OP NEEDS TO GIVE IT A TRY HIMSELF!
--------------------
mr bob, just try what arildno suggested you.
2^{x}=e^{xlog(2)}
Now differentiate both sides with respect to x, we have:
(2^{x})'=(e^{xlog(2)})' = ?
--------------------
Or one can try another way:
Let:
y = 2 ^ x
Take log of both sides:
\ln y = \ln (2 ^ x) = x \ln 2
Now differentiate both sides wth respect to x gives:
(\ln y)' = (x \ln 2)'
\Leftrightarrow \frac{y'_x}{y} = \ln 2
\Leftrightarrow y'_x = y \ln 2 = 2 ^ x \star ln 2
Now let's try the first way to see if you arrive at the same answer.
From there, can you solve your problem? :)

 

[WARGREYMONKKTL]

that is wrong!

y = a^u

y' = lna * a^u * u'

y = 2^x y' = ln2 * 2^x * 1
derivative of (x-4) is 1

so the answer is y' = 2^x * ln2 + 1

sorry to say so but you see the obvious right there!
please correct. thanks

 

[d_leet]

sorry to say so but you see the obvious right there!
please correct. thanks

What are you talking about? I can't see anything wrong with that...

 

[WARGREYMONKKTL]

sorry because i don't understand the term "+1" in the result y'
please explain .

 

[d_leet]

sorry because i don't understand the term "+1" in the result y'
please explain .

Did you look at the original post? Look at the funtion that we are trying to differentiate here and maybe you'll figure it out.

 

[WARGREYMONKKTL]

ok i agree with you!
sorry aabout that.
i want to ask a question if you can solve?
y= ksin2x.tan3x all over 3cos4x with x is an accute angle and k is a constant.
can you have the general derivative and the value of that function at the point which is the inteception of y with y =-x and the value of k is 1:)
this is not a challenge but a hard question to ask you.
this is my apology
see ya!

 

[d_leet]

ok i agree with you!
sorry aabout that.
i want to ask a question if you can solve?
y= ksin2x.tan3x all over 3cos4x with x is an accute angle and k is a constant.
can you have the general derivative and the value of that function at the point which is the inteception of y with y =-x and the value of k is 1:)
this is not a challenge but a hard question to ask you.
this is my apology
see ya!

I'm sorry, but I can't make heads or tails of what you just said, I get the part where you say what the function is but what is that about a general derivative and a point which is an intersection of something?

 

[HallsofIvy]

ok i agree with you!
sorry aabout that.
i want to ask a question if you can solve?
y= ksin2x.tan3x all over 3cos4x with x is an accute angle and k is a constant.
can you have the general derivative and the value of that function at the point which is the inteception of y with y =-x and the value of k is 1:)
this is not a challenge but a hard question to ask you.
this is my apology
see ya!

First, it's a really bad idea to "hijack" someone elses thread to ask your onw question- start your own thread. Second, even without using LaTex you ought to be able to write that more clearly. I think you mean
y= ksin(2x)tan(3x)/(3cos(4x)).

If you know anything at all about calculus you should see that neither k nor the fact that "x is an acute angle" are relevant. Yes, it certainly has a "general derivative"- its tedious but just apply the product, quotient, and chain rules.

As for solving x= sin(2x)tan(3x)/(3cos(4x)), that's much harder. In general equations like that cannot be solved algebraically. I would recommend using Newton's method to get a numerical solution.

 

[WARGREYMONKKTL]

i am sorry.
let take my apology.
actually i just plan to play with you.
i am a jounor in high school.
i don't want this will be a mathematical fight between us.
please forgive me for that.
can you give me the result for the Newton's method. i really don't know about that method.
thanks!

 

[VietDao29]

i am sorry.
let take my apology.
actually i just plan to play with you.
i am a jounor in high school.
i don't want this will be a mathematical fight between us.
please forgive me for that.
can you give me the result for the Newton's method. i really don't know about that method.
thanks!

Here's an article on Newton's method.
It state that:
x_{n + 1} = x_n - \frac{f(x_n)}{f'(x_n)}
Now the solution to the equation f(x) = 0 is given by:
x = \lim_{n \rightarrow \infty} x_n
There's also an example in the article, just apply it here, and see what you get.
Can you go from here? :)