can someone please help me with this problem....
.e^(x/2)
----------
..x^(5)
can someone please give me an idea of how to do this problem... thank you!!
Lets assume that you are asking about the derivative of:Originally Posted by demi
.Well this can be done in a number of ways, you could use the quotient rule,
or the product rule to find this derivative. Now I don't have that good a
memory, and so never bother with the quotient rule, always using the product rule:
.
Here we let:
,
and:
.
Now:
,
and:
.
So now you have enough to assemble the derivative
RonL
Differentiation with respect to x.Originally Posted by demi
d/dx [u/v] = [v*du/dx -u*dv/dx] / v^2 ------------***
d/dx [e^(x/2) / x^5]
= {[(x^5)*e^(x/2) *(1/2)] -[e^(x/2) *(5x^4)]} / [(x^5)^2]
= {(1/2)(x^5)e^(x/2) -5(x^4)e^(x/2)} / x^10
= [e^(x/2)][(x^5 /2) -(5x^4)]/ x^10
= [e^(x/2)][1 /(2x^5) -5/(x^6)] ---------------------(i)
= [e^(x/2)][{[1 /(2x^5)]*(x/x)} -{[5/(x^6)]*(2/2)]}
= [e^(x/2)][x /(2x^6) -10/(2x^6)]
= [e^(x/2)][(x -10) /(2x^6)]
= (1/2)[e^(x/2)][(x -10) /(x^6)] ------------answer.
--------------------------------------
d/dx [u*v] = v*du/dx +u*dv/dx ----------***
d/dx [e^(x/2) / x^5]
= d/dx [(x^(-5))(e^(x/2))]
= e^(x/2) *[-5 x^(-6)] +[x^(-5) *[e^(x/2) *(1/2)]
= [e^(x/2)][-5/(x^6) +(1 /(2x^5)]
= [e^(x/2)][1/(2x^5) -5/(x^6)] ------------same as (i) above.
Etc...