How do find the derivative using first principles formula?

#4x##sqrtx#

#4x##sqrtx#

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Feb 22, 2018

Answer:

Derivative is #6sqrtx#

Explanation:

In first principle method we calculate derivative for #y=f(x)#, using definition of derivative as

#(dy)/(dx)# or #(df)/(dx)=lim_(h->0)(f(x+h)-f(x))/h#

Here as #f(x)=4xsqrtx#, #f(x+h)=4(x+h)sqrt(x+h)#

and #(df)/(dx)=lim_(h->0)(4(x+h)sqrt(x+h)-4xsqrtx)/h#

= #4lim_(h->0)(((x+h)sqrt(x+h)-xsqrtx)((x+h)sqrt(x+h)+xsqrtx))/(h((x+h)sqrt(x+h)+xsqrtx))#

= #4lim_(h->0)((x+h)^3-x^3)/(h((x+h)sqrt(x+h)+xsqrtx))#

= #4lim_(h->0)(x^3+3hx^2+3h^2x+h^3-x^3)/(h((x+h)sqrt(x+h)+xsqrtx))#

= #4lim_(h->0)(3x^2+3hx+h^2)/((x+h)sqrt(x+h)+xsqrtx)#

= #4xx(3x^2)/(2xsqrtx)#

= #2xx(3x^2)/x^(3/2)#

= #6sqrtx#

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