If #f(x)= 2 x^2 + x # and #g(x) = sqrtx + 1 #, how do you differentiate #f(g(x)) # using the chain rule?

1 Answer
Feb 23, 2016

Answer:

#f'(g(x)) = 2 +5/(2sqrtx)#

Explanation:

Find f(g(x)) # = f(sqrtx + 1 )#

substitute #x = sqrtx + 1 " in f(x) to obtain "#

f(g(x)) = #2(sqrtx + 1 )^2 + (sqrtx + 1 )#

distribute #(sqrtx + 1 )^2 = x + 2sqrtx +1 #

#rArr f(g(x)) = 2(x + 2sqrtx +1) + sqrtx + 1 #

# = 2x + 4sqrtx + 2 + sqrtx + 1 = 2x + 5sqrtx + 3 #

and writing #2x + 5sqrtx + 3" as " 2x + 5x^(1/2) + 3 #

now differentiate using#color(blue)" power rule "#

ie #d/dx(ax^n) = nax^(n-1) " term by term "#

#rArr f'(g(x)) = 2 + 5/2 x^(-1/2) = 2 + 5/(2sqrtx) #