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You entered:

exp(c1+c2*x)^(-5x)

What you mean?

Derivative of exp(c1+c2*x)^(-5x)

Function f() - derivative -N order at the point
v

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Piecewise:

The solution

You have entered [src]
            -5*x
/ c1 + c2*x\    
\e         /    
(ec1+c2x)5x\left(e^{c_{1} + c_{2} x}\right)^{- 5 x}
Detail solution
  1. Don't know the steps in finding this derivative.

    But the derivative is

    (5x)5x(log(5x)+1)\left(- 5 x\right)^{- 5 x} \left(\log{\left(- 5 x \right)} + 1\right)


The answer is:

(5x)5x(log(5x)+1)\left(- 5 x\right)^{- 5 x} \left(\log{\left(- 5 x \right)} + 1\right)

The first derivative [src]
            -5*x                                                      
/ c1 + c2*x\     /       / c1 + c2*x\           c1 + c2*x  -c1 - c2*x\
\e         /    *\- 5*log\e         / - 5*c2*x*e         *e          /
(5c2xec1c2xec1+c2x5log(ec1+c2x))(ec1+c2x)5x\left(- 5 c_{2} x e^{- c_{1} - c_{2} x} e^{c_{1} + c_{2} x} - 5 \log{\left(e^{c_{1} + c_{2} x} \right)}\right) \left(e^{c_{1} + c_{2} x}\right)^{- 5 x}
The second derivative [src]
              -5*x /                                  2\
  / c1 + c2*x\     |          /          / c1 + c2*x\\ |
5*\e         /    *\-2*c2 + 5*\c2*x + log\e         // /
5(2c2+5(c2x+log(ec1+c2x))2)(ec1+c2x)5x5 \left(- 2 c_{2} + 5 \left(c_{2} x + \log{\left(e^{c_{1} + c_{2} x} \right)}\right)^{2}\right) \left(e^{c_{1} + c_{2} x}\right)^{- 5 x}
The third derivative [src]
               -5*x /                            2       \                         
   / c1 + c2*x\     |    /          / c1 + c2*x\\        | /          / c1 + c2*x\\
25*\e         /    *\- 5*\c2*x + log\e         //  + 6*c2/*\c2*x + log\e         //
25(6c25(c2x+log(ec1+c2x))2)(c2x+log(ec1+c2x))(ec1+c2x)5x25 \left(6 c_{2} - 5 \left(c_{2} x + \log{\left(e^{c_{1} + c_{2} x} \right)}\right)^{2}\right) \left(c_{2} x + \log{\left(e^{c_{1} + c_{2} x} \right)}\right) \left(e^{c_{1} + c_{2} x}\right)^{- 5 x}