Math Differentiation Problem -- differentiating powers?

Question

Why does

d/dx c^(1-x) = -c^(1-x)log(c)?

where c is a constant.

I would like to know how to differentiate with respect to x when x is a power.

Question#061205120802

Answer 1

From the definition, d/dt (ln t) = 1/t, you can derive d/dt (e^t) = e^t, using implicit differentiation.

By the chain rule, d/dt (e^f(t)) = e^(f(t)) * f'(t).
Note that c^x = e^(ln(c)*x), since e^(ln(c)) = c.
Then d/dt (c^t) = d/dt (e^(ln(c)*t) = e^(ln(c)*t) * ln(c) = c^t * ln(c), since d/dt (ln(c)*t) = ln(c).

Another application of the chain rule gives
d/dx (c^(1-x)) = c^(1-x) * ln(c) * (-1), since d/dx (1-x) = -1.

Answer 2

Let y= a^u , where u is any function of x
Then ln(y) = ln(u^x) = u ln(a)

Then (1/y) dy/dx = ln(a)du/dx
dy/dx = ln(a) * y * du/dx
dy/dx = ln (a) * a^u *du/dx
In your case u = (1-x) and a = c
Thus du/dx = -1
Plug the above in and get:
dy/dx = -ln(c) * c ^(1-x)

Note that we are using logs to the base e here so we use ln instead of log. Note also that a=c >1.

See if you can show that if y= 2^(2x-1), then
dy/dx = (ln 2) *2^2x
[Hint: 2^(2x-1) * 2 = 2^2x

Hoped this helped.

Answer 3

as c can be written c= e^ln c (definition of the ln of a number)

you can rewrite c^(1-x) as (e ^lnc )^(1-x)

and (e^x)^y = e^xy

so c^(1-x) = e^(ln c(1-x)) = e^(lnc-xlnc)

form e^(u) with the derivate e^(u) *du/dx

du/dx =-lnc

and you find the result

Answer 4

One side of the equation is x^y and the other x^-y.
Y = what you powering x by.

So x^6 would be coverted to x^-6

Answer 5

logs .... allow us to transform a exponential equation into an algebraic one ....

the derivative of e^(U) is .....e^U*du