Use the limit definition to find the derivative of f(x)=3/sqrtx at the number a, and use it to find the equation of the tangent line to the curve at the point (1,3)?

f(x)= #3/sqrtx#

1 Answer
Steve M
Mar 6, 2018

# f'(a) = (-3)/(a(2sqrt(a)) )#

# y = -3/2 x + 9/2 #

Explanation:

Using the limit definition of the derivative, we have:

# f'(a) =lim_(h rarr 0) (f(a+h)-f(a))/h #

# \ \ \ \ \ \ \ \ \ =lim_(h rarr 0) (3/sqrt(a+h)-3/sqrt(a))/h #

# \ \ \ \ \ \ \ \ \ =lim_(h rarr 0) ((3sqrt(a) - 3sqrt(a+h))/(sqrt(a)sqrt(a+h) ) )/h #

# \ \ \ \ \ \ \ \ \ =lim_(h rarr 0) (3(sqrt(a) - sqrt(a+h)))/(hsqrt(a)sqrt(a+h) ) #

# \ \ \ \ \ \ \ \ \ =lim_(h rarr 0) (3(sqrt(a) - sqrt(a+h)))/(hsqrt(a)sqrt(a+h) ) * ((sqrt(a) + sqrt(a+h)))/((sqrt(a) + sqrt(a+h)))#

# \ \ \ \ \ \ \ \ \ =lim_(h rarr 0) (3((a) - (a+h)))/(hsqrt(a)sqrt(a+h)(sqrt(a) + sqrt(a+h)) )#

# \ \ \ \ \ \ \ \ \ =lim_(h rarr 0) (3(-h))/(hsqrt(a)sqrt(a+h)(sqrt(a) + sqrt(a+h)) )#

# \ \ \ \ \ \ \ \ \ =lim_(h rarr 0) (-3)/(sqrt(a)sqrt(a+h)(sqrt(a) + sqrt(a+h)) )#

# \ \ \ \ \ \ \ \ \ = (-3)/(sqrt(a)sqrt(a)(sqrt(a) + sqrt(a)) )#

# \ \ \ \ \ \ \ \ \ = (-3)/(a(2sqrt(a)) )#

Which is the gradient of the tangent at the point #x=a#, And so having verified that #(1,3)# lies on the given curve we can form the tangent equation using the point-slope equation of a straight line:

# y-y_1= m(x-x_1) #

And we get the equation:

# y - 3 = (-3)/((2sqrt(1)) ) (x - 1 ) #
# :. y - 3 = (-3/2) (x - 1 ) #
# :. y - 3 = -3/2 x + 3/2 #
# :. y = -3/2 x + 9/2 #

Which we can verify graphically:

graph{(y-3/sqrt(x))(y +3/2 x - 9/2)=0 [-5, 10, -5, 10]}

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