Question 202999
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{{{f(x)=8e^(-7^x)}}} 

Let {{{y = f(x)}}}

{{{y=8e^(-7^x)}}} 

Take natural logs of both sides:

{{{ln(y)=ln(8e^(-7^x))}}}

Use the fact that {{{ln(uv)=ln(u)+ln(v)}}}

{{{ln(y)=ln(8) + ln(e^(-7^x)))}}} 

Use the fact that {{{ln(e^u) = u}}} to rewrite the last term:

{{{ln(y)=ln(8) + (-7^x)}}}

{{{ln(y)=ln(8) -7^x}}}

Now finally we can take the derivative implicitly.  

1. We use {{{d/dx}}}{{{ln(u)="u'"/u}}} to take the 
derivative of the term on the left.

2. {{{ln(8)}}} is a constant so its derivative is 0.

3. We use {{{d/dx}}}{{{a^u=a^u*ln(a)*"u'"}}} for the last term

So for {{{ln(y)=ln(8) -7^x}}} we take the derivatives:

{{{"y'"/y=0 -7^x*ln(7)*1}}}

{{{"y'"/y=-7^x*ln(7)}}}

Multiply both sides by {{{y}}}

{{{"y'"=-7^x*ln(7)*y}}}

Use {{{y=8e^(-7^x)}}} to replace {{{y}}}

{{{"y'"=-7^x*ln(7)*(8e^(-7^x))}}}

Rearranging the factors:

{{{"y'"=-8*7^x*e^(-7^x)*ln(7)}}}

Replace {{{y}}} by {{{"f'(x)"}}}

{{{"f'(x)"=-8*7^x*e^(-7^x)*ln(7)}}}

Edwin</pre>