You need to use limit definition of derivatives such that:

`f'(x) = lim_(Delta x-gt 0) (f(x+Delta x) - f(x))/(Delta x)`

`f'(x) = lim_(Delta x-gt 0) (2^(5(x+Delta x)) - 2^(5x))/(Delta x)`

`f'(x) = lim_(Delta x-gt 0) (2^(5x)*2^(Delta x) - 2^(5x))/(Delta x)`

You need to factor out `2^(5x)` yields:

`f'(x) = lim_(Delta x-gt 0) (2^(5x)*(2^(Delta x) - 1))/(Delta x)`

You need to consider `2^(5x)` as constant such that:

`f'(x) = 2^(5x)*lim_(Delta x-gt 0) (2^(Delta x) - 1))/(Delta x)`

You may use the formula of special limit such that:

`lim_(x-gt 0) (a^x - 1)/x = ln a`

`lim_(Delta x-gt 0) (2^(Delta x) - 1)/(Delta x) = ln 2`

`f'(x) = 2^(5x)*5*ln 2`

`f'(x) = 2^(5x)*ln 2^5 =gt f'(x) = 2^(5x)*ln 32`

You should evaluate the slope of tangent line at the curve at point x=1 such that:

`f'(1) = 2^5*ln 32 =gt f'(1) = 32*ln 32`

**Hence, evaluating the slope of tangent line at the curve at point x=1 yields `f'(1) = 32*ln 32` .**

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