Find the derivative using differentiation?

d/dx = x^3+2

We have three rules to use:

d/dx ax^n = nax^n-1

d/dx ax = a

and d/dx c = 0 where c in a constant.

So our three parts are 1/4*x^4, 2x and 2. We can differentiate them separately then add them together to get the derivative.

Part 1: d/dx 1/4*x^4 = 4*1/4*x^(4-1) = x^3

Part 2: d/dx 2x = 2

Part 2: d/dx 2 = 0

So we have dy/dx = x^3+2+0 or just x^3+2

The same as any other algebraic differentiation

x^4/4 + 2 x + 2 =

X^3 + 2

f ` (x) = x ³ + 2

((1/4)*x^4))+2x+2=

4*(1/4)*x^(4-1)+1*2x^(1-1)+2*0 = x^3+2x

You take the power of the variable in front of the variable, then you subtract 1 from the exponent.i.e you decrease the exponent to its next low level. I mean if the exponent is 3 it has to decrease to 2, while 3 multiply the variable in front. The constant numbers will disappear, i.e all terms without a variable. In your case, +2 is a constant then it become 0.

logarithmic differentiation is a pretty long process, but.... here are the steps: 1) take the log of both sides logy = log(1 + etc) 2) drop the exponent logy = (sinx)log(1 + 2^x) 3) derive each side --- use product rule on the right side! (1/y) y' = sinx (1/(1+2^x) (2^x)(ln2) + log(1 + 2^x)(cosx) 4) times both sides by y (the original function) to get a final answer for y' 5) y' = [sinx (1/(1+2^x) (2^x)(ln2) + log(1 + 2^x)(cosx)] ((1 + 2^x)^sinx I did this pretty fast so i hope i didn't make any silly mistakes but ...those are the steps!

(1/4)x^4 has a derivative of x^3, 2x has a derivative of 2, and 2 has a derivative of 0, so add them up to get

x^3 + 2.

Remember that the derivative of x^n is n*x^(n-1)