# Find the derivative of f(x)= x^2 + 5^x using lim h-->0 f(x+h)-f(x)/h You need to use limit definition of derivatives such that:

`lim_(h-gt0) ((x+h)^2 + 5^(x+h) - x^2 - 5^x)/h`

You need to form two groups of terms such that:

`lim_(h-gt0) ((x+h)^2 - x^2)/h + lim_(h-gt0)(5^(x+h) -5^x)/h`

You should identify the difference of squares `(x+h)^2 - x^2 = (x + h - x)(x +...

You need to use limit definition of derivatives such that:

`lim_(h-gt0) ((x+h)^2 + 5^(x+h) - x^2 - 5^x)/h`

You need to form two groups of terms such that:

`lim_(h-gt0) ((x+h)^2 - x^2)/h + lim_(h-gt0)(5^(x+h) -5^x)/h`

You should identify the difference of squares `(x+h)^2 - x^2 = (x + h - x)(x + h + x)` `(x+h)^2 - x^2 = h(2x+h) `

`lim_(h-gt0) ((x+h)^2 - x^2)/h = lim_(h-gt0) (h(2x+h) )/h `

`lim_(h-gt0) (h(2x+h) )/h =lim_(h-gt0) (2x+h) = 2x + 0 = 2x`

You need to evaluate the limit `lim_(h-gt0) (5^(x+h) - 5^x)/h`  such that:

`lim_(h-gt0) (5^x)(5^(x+h-x) - 1)/h = lim_(h-gt0) (5^x)(5^h -1)/h`

You need to use the special limit `lim_(x-gt0) (a^x - 1)/x = ln a`  such that:

`lim_(h-gt0) (5^h -1)/h = ln 5`

`lim_(h-gt0) (5^x)(5^h -1)/h = 5^x*ln 5`

Adding the results of limits `lim_(h-gt0) ((x+h)^2 - x^2)/h`  and `lim_(h-gt0) (5^(x+h) - 5^x)/h`  yields:

`lim_(h-gt0) ((x+h)^2 + 5^(x+h) - x^2 - 5^x)/h = 2x + 5^x*ln 5`

Hence, evaluating the limit to the function using derivative definition yields`lim_(h-gt0) ((x+h)^2 + 5^(x+h) - x^2 - 5^x)/h = 2x + 5^x*ln 5.`

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