Find the slope of the tangent to the curve y=1x at the point where x = a

As per the formula #2 from this section:f′(x)=limh→0f(x+h)−f(x)hUse f(x)=1xf′(x)=limh→01x+h−1xhMultiply both the numerator and the denomiantor by x+hxf′(x)=limh→0x−x+hhx+hx}Multiply both the numerator and the denomiantor by x+x+hf′(x)=limh→0x−x+hhx+hx⋅x+x+hx+x+hRecall that: A2−B2=(A−B)(A+B)f′(x)=limh→0x−(x+h)hx+hx(x+x+h}Cancel both the numerator and the denominatorf′(x)=limh→0−1x+hx(x+x+h)Note that substituting h=0 no longer makes the denominator 0, so we can evaluate this limit by direct substitution.f′(x)=−1x+0x(x+x+0)=−12xRecall that: f'(a) is the slope of the tangent of y=f(x) at x=aResult: Slope of the tangent at x=a is 12aa

"Find the slope of the tangent to the..." was answered by our community

Marvin Mccormick

Answered question

2021-10-02

Find the slope of the tangent to the curve

y = \frac{1}{\sqrt{x}}

at the point where x = a

Answer & Explanation

averes8

Skilled2021-10-03Added 92 answers

As per the formula #2 from this section:

f^{'} (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}

Use

f (x) = \frac{1}{\sqrt{x}}

f^{'} (x) = lim_{h \to 0} \frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h}

Multiply both the numerator and the denomiantor by

\sqrt{x + h} \sqrt{x}

f^{'} (x) = lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h \sqrt{x + h} \sqrt{x}}}

Multiply both the numerator and the denomiantor by

\sqrt{x} + \sqrt{x + h}

f^{'} (x) = lim_{h \to 0} \frac{\sqrt{x} - \sqrt{x + h}}{h \sqrt{x + h} \sqrt{x}} \cdot \frac{\sqrt{x} + \sqrt{x + h}}{\sqrt{x} + \sqrt{x + h}}

Recall that:

A^{2} - B^{2} = (A - B) (A + B)

f^{'} (x) = lim_{h \to 0} \frac{x - (x + h)}{h \sqrt{x + h} \sqrt{x} (\sqrt{x} + \sqrt{x + h}}}

Cancel both the numerator and the denominator

f^{'} (x) = lim_{h \to 0} \frac{- 1}{\sqrt{x + h} \sqrt{x} (\sqrt{x} + \sqrt{x + h})}

Note that substituting h=0 no longer makes the denominator 0, so we can evaluate this limit by direct substitution.

f^{'} (x) = \frac{- 1}{\sqrt{x + 0} \sqrt{x} (\sqrt{x} + \sqrt{x + 0})} = - \frac{1}{2 \sqrt{x}}

Recall that: f'(a) is the slope of the tangent of

y = f (x)

at x=a
Result: Slope of the tangent at

x = a

\frac{1}{2 a \sqrt{a}}

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Algebra II

Didn't find what you were looking for?