Differentiating Complicated Functions

Formulas with Ln are:
Ln(x) = ( x' ) / x

Ln(x*y) = Lnx + Lny = (x' / x) + (y' / y)
#2
the derivative of 2^x=(2^x) *ln 2 by the formula d/dx (a^x) = a^x *lna

Ln(x*y) --> ln[(x^2)(2^x)] = ln(x^2)+Ln(2^x) = 2x/x^2 + (2^x) *ln 2

the calc can do ln2, simplify to 2/x + .69x

#3
csc(5x)-cot(5x) is one function so Ln(csc(5x)-cot(5x)) =

[derivative of csc(5x)-cot(5x)] / csc(5x)-cot(5x) Just plow through it.
-5csc(5x)*cot(5x)+5csc^2(5x) / csc(5x)-cot(5x)

the first one is done in parts, (x^1/2)+(x^-4/3)+(2^x)-(x^pi)+(e^2)
x^1/2 = 1/2sqrt(x)
(x^-4/3) = 1/3 * f ' x^-4 = 1/3 * -4x^-5
(2^x) = (2^x) *ln 2
(x^pi) = pi*x^2.14
e^2 is a constant so its derivative is 0

Find you teacher b/c catching up is the best resource.

Wish I had a link for you, but I'll explain, and maybe someone else will post a link.

The big key your missing here is d/dx(e^x) = e^x and d/dx(ln x) = 1/x. You may have missed that class.

So, one the first one:

y = (x)+(x^-4/3)+(2^x)-(x^pi)+(e^2)

dy/dx = 1 -(4/3)x^(-7/3) + d/dx(2^x) - d/dx(x^π) + d/dx(e²)

So let's look at those last three terms:

d/dx(e²): Don't let the e fool you, e is a constant, so e² is a constant so d/dx(e²) = 0

d/dx(x^π): Don't worry about π. It's a constant too, just like any other real number so treat like you would any other power. So d/dx(x^π) = πx^(π-1).

d/dx(2^x): This is the tricky one. Let's work it out:

y = 2^x
ln y = x ln 2

So to differentiate, use the Chain Rule on ln y. Remember, d/dx(ln x) = 1/x

(1/y)*dy/dx = d/dx(x ln 2) = ln 2

Now solve for dy/dx:

dy/dx = y ln 2 = (2^x)(ln 2)

Memorize this formula: d/dx(a^x) = (a^x)(ln a)

So let's substitute now into the original equation where we left off:

dy/dx = 1 - (4/3)x^(-7/3) +(2^x)(ln 2) - πx^(π-1) + 0

And you simplify from there. On the last two use your new knowledge:

d/dx(ln x) = 1/x
d/dx(a^x) = (a^x)(ln a)

Along with the Chain Rule and you should be able to work it out.