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)?

Question

f(x)= #3/sqrtx#

Charles Anctil · Accepted Answer

# f'(a) = (-3)/(a(2sqrt(a)) )#
# y = -3/2 x + 9/2 #

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]}

Christopher Allers · Answer

undefined To find the derivative of $f(x) = \frac{3}{\sqrt{x}}$ using the limit definition, we'll use the formula:

$$ f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} $$

First, let's find $f(a + h)$:

$$ f(a + h) = \frac{3}{\sqrt{a + h}} $$

Next, let's find $f(a)$:

$$ f(a) = \frac{3}{\sqrt{a}} $$

Now, let's plug these into the formula:

$$ f'(a) = \lim_{h \to 0} \frac{\frac{3}{\sqrt{a + h}} - \frac{3}{\sqrt{a}}}{h} $$

To simplify this expression, we'll rationalize the numerator:

$$ f'(a) = \lim_{h \to 0} \frac{3\sqrt{a} - 3\sqrt{a + h}}{h\sqrt{a}\sqrt{a + h}} $$

$$ f'(a) = \lim_{h \to 0} \frac{3\sqrt{a} - 3\sqrt{a + h}}{h\sqrt{a}\sqrt{a + h}} \cdot \frac{\sqrt{a} + \sqrt{a + h}}{\sqrt{a} + \sqrt{a + h}} $$

$$ f'(a) = \lim_{h \to 0} \frac{3a - 3(a + h)}{h\sqrt{a}(a + h)} $$

$$ f'(a) = \lim_{h \to 0} \frac{-3h}{h\sqrt{a}(a + h)} $$

$$ f'(a) = \lim_{h \to 0} \frac{-3}{\sqrt{a}(a + h)} $$

$$ f'(a) = \frac{-3}{\sqrt{a} \cdot a} $$

Now, to find the equation of the tangent line at the point (1, 3), we need to find the slope of the tangent line at $x = 1$.

$$ f'(1) = \frac{-3}{\sqrt{1} \cdot 1} = -3 $$

So, the slope of the tangent line is $m = -3$.

Using the point-slope form of the equation of a line, $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the point (1, 3), we plug in the values:

$$ y - 3 = -3(x - 1) $$

Now, we can simplify this equation to get the equation of the tangent line.