How do you determine the instantaneous rate of change of #y(x) = sqrt(3x + 1)# for #x = 1#?

Question

How do you determine the instantaneous rate of change of #y(x) = sqrt(3x + 1)# for  #x = 1#?

Cameron Alexander · Accepted Answer

Apply the chain rule for derivatives and substitute #x=1# in the result. #(d (y(x)))/(dx) = (1/2)(3x+1)^(-1/2)*(3)# #(d (y(1)))/(dx) = 3/(2 sqrt(3+1)) = 3/4# #(d (y(1)))/(dx)# is the instantaneous rate of change of #y(x)# at #x=1#

Christopher Agresta · Answer

The specifics will depend on your textbook and the definition selected by your instructor if you have to work from it:

One will be available to you:

#lim_(xrarr1)(y(x)-y(1))/(x-1)# or #lim_(hrarr0)(y(1+h)-y(1))/h#

First Definition:

#lim_(xrarr1)(y(x)-y(1))/(x-1)=lim_(xrarr1)(sqrt(3x+1)-sqrt(3(1)+1))/(x-1)#

#=lim_(xrarr1)(sqrt(3x+1)-2)/(x-1)#

Notice that substitution leads to indeterminate form: #0/0#. The trick (technique) to try in this case is to rationalize the numerator. Multiply numerator and denominator by the conjugate of the numerator. (Multiply the top and bottom by #(sqrt(3x+1) + 2)#

#lim_(xrarr1)(y(x)-y(1))/(x-1)=lim_(xrarr1)[((sqrt(3x+1)-2))/(x-1) ((sqrt(3x+1) + 2))/((sqrt(3x+1) + 2))]#

#=lim_(xrarr1)(sqrt(3x+1)^2-2^2)/((x-1)(sqrt(3x+1) + 2)#

#=lim_(xrarr1)(3x+1-4)/((x-1)(sqrt(3x+1) + 2)#

#=lim_(xrarr1)(3x-3)/((x-1)(sqrt(3x+1) + 2)#

#=lim_(xrarr1)(3(x-1))/((x-1)(sqrt(3x+1) + 2)# (Still form #0/0#)

#=lim_(xrarr1)3/(sqrt(3x+1) + 2)# (No longer form #0/0#)

#=3/(sqrt(3(1)+1) + 2)=3/(2+2)=3/4#

Second Definition:

#lim_(hrarr0)(y(1+h)-y(1))/h =lim_(hrarr0)(sqrt(3(1+h)+1)-sqrt(3(1)+1))/h#

#=lim_(hrarr0)(sqrt(3h+4)-2)/h#

Notice that substitution leads to indeterminate form: #0/0#. The trick (technique) to try in this case is to rationalize the numerator. Multiply numerator and denominator by the conjugate of the numerator. (Multiply the top and bottom by #(sqrt(3h+4) + 2)#

#lim_(hrarr0)(y(1+h)-y(1))/h =lim_(hrarr0)((sqrt(3h+4)-2))/h ((sqrt(3h+4) + 2))/((sqrt(3h+4) + 2))#

# =lim_(hrarr0)((sqrt(3h+4)^2-2^2))/(h(sqrt(3h+4) + 2))#

# =lim_(hrarr0)(3h+4-4)/(h(sqrt(3h+4) + 2))# , (still form #0/0#)

# =lim_(hrarr0)(3h)/(h(sqrt(h+4) + 2))# , (still form #0/0#)

# =lim_(hrarr0)3/(sqrt(h+4) + 2)# , (no longer form #0/0#)

#=3/(sqrt(0+4) + 2)=3/(2+2)=3/4#

William Abdalla · Answer

undefined To determine the instantaneous rate of change of $ y(x) = \sqrt{3x + 1} $ at $ x = 1 $, you would first find the derivative of the function with respect to $ x $ using the power rule. Then, evaluate the derivative at $ x = 1 $. The derivative of $ y(x) $ is $ y'(x) = \frac{3}{2\sqrt{3x + 1}} $. Plugging in $ x = 1 $, we get $ y'(1) = \frac{3}{2\sqrt{3(1) + 1}} = \frac{3}{2\sqrt{4}} = \frac{3}{4} $. Therefore, the instantaneous rate of change of $ y(x) $ at $ x = 1 $ is $ \frac{3}{4} $.