f'(x)=lim h-->0 f(x+h)-f(x)/h solve:  f(x)= sq rt. (5x+3)Use 4 step process to find the derivative....

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You need to evaluate derivative of the function f'(x) such that:

`f'(x) = lim_(h-gt0) (sqrt(5x + 5h + 3) - sqrt(5x+3))/h`

You need to multiply fraction by conjugate of `(sqrt(5x + 5h + 3) - sqrt(5x+3))`  such that:

`f'(x) = lim_(h-gt0) ((sqrt(5x + 5h + 3) - sqrt(5x+3))(sqrt(5x + 5h + 3)+ sqrt(5x+3)))/(h((sqrt(5x + 5h + 3)+ sqrt(5x+3)) ))`

You need to use the special product `a^2 - b^2 = (a-b)(a+b)`

`((sqrt(5x + 5h + 3) - sqrt(5x+3))(sqrt(5x + 5h + 3) + sqrt(5x+3))) = 5x + 5h + 3 - 5x - 3`

Reducing like terms yields:

`((sqrt(5x + 5h + 3) - sqrt(5x+3))(sqrt(5x + 5h + 3) + sqrt(5x+3))) = 5h`

`f'(x)= lim_(h-gt0) (5h)/(h(sqrt(5x + 5h + 3) + sqrt(5x+3))) `

`f'(x)= lim_(h-gt0) 5/(sqrt(5x + 5h + 3) + sqrt(5x+3))`

`f'(x)= 5/(2sqrt(5x+3))`

Hence, evaluating the first derivative of function using first principle yields `f'(x)= 5/(2sqrt(5x+3)).`

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