Learning Calculus: Chain Rule and Derivatives

[MotoPayton]

I am currently learning calculus and just had my lecture on the chain rule.

I noticed that we haven't learned how to take the derivative of a function like 2^2+x or 3^4+x.
Any example works.. Is this something I will learn later as I progress through calculus or what?

 

[Mark44]

I am currently learning calculus and just had my lecture on the chain rule.

I noticed that we haven't learned how to take the derivative of a function like 2^2+x or 3^4+x.
Any example works.. Is this something I will learn later as I progress through calculus or what?

Assuming you meant what you wrote, these are very simple to differentiate. 2^2 + x = 4 + x, and its derivative is 1.

3^4 + x = 81 + x, and its derivative is also 1.

Now, assuming you meant 2^(2 + x) = 2^{2 + x}, and 3^(4 + x) = 3^{4 + x}, these functions can be differentiated when you know the chain rule form of the derivative of the exponential function.

 

[mynameisfunk]

chain rule : F'(x) = f(g(x)) g'(x)

example: take f(x) = 2^sin(x) .....f'(x)=2^sin(x)cos(x)

 

[phyzguy]

Not quite right. The derivative of 2^x is not just 2^x, you need to proceed as follows:
[tex]\frac{d}{dx} 2^x = \frac{d}{dx} e^{x log2} = \frac{d}{dx} (e^x)^{log2} = e^x log2*(e^x)^{log2-1} = log2*2^x [/tex]

So:
[tex]\frac{d}{dx} 2^{sin(x)} = log2*cos(x)*2^{sin(x)}[/tex]

 

[mynameisfunk]

my bad. not even a good example.

 

[HallsofIvy]

Generally, for any positive a, the derivative of [itex]a^x[/itex], with respect to x, is [itex](ln a)a^x[/itex]. Of course, if a= e, ln(a)= ln(e)= 1.

 

[paulfr]

So finish this off ...
Derivative of 2^(2 +x) = (Ln 2) [2^(2+x)] (2+x)' = (Ln 2) [2^(2+x)]
and
Derivative of 3^(4 +x) = (Ln 3) [3^(4+x)] (4+x)' = (Ln 3) [3^(4+x)]

Just for fun checkout Derivative of 2^(2+x^2)
http://www.wolframalpha.com/input/?i=derivative+2^%282+%2B+x^2%29+dx
Click on "Show Steps" to see the solution