# Tangent to the curve

• Feb 9th 2009, 12:48 PM
Lipticboven
Tangent to the curve
I need help finding the tangent to the curve $\displaystyle y = \frac{3}{1 + \sqrt{x}}$, at the point (4,1). Im really only having problems finding the gradent of the tangent, i already know the answer but everytime i try i either get the gradent as 0 or infinity which isnt correct.

Im not ment to be using differentiation, but instead using limits.
• Feb 11th 2009, 09:12 AM
Moo
Hello,
Quote:

Originally Posted by Lipticboven
I need help finding the tangent to the curve $\displaystyle y = \frac{3}{1 + \sqrt{x}}$, at the point (4,1). Im really only having problems finding the gradent of the tangent, i already know the answer but everytime i try i either get the gradent as 0 or infinity which isnt correct.

Im not ment to be using differentiation, but instead using limits.

Let $\displaystyle f(x)=\frac{3}{1+\sqrt{x}}$

You have to find $\displaystyle f'(4)$ in order to find the gradient.

So since you have to use limits and not differentiation, remember that :
$\displaystyle f'(a)=\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$

So here :
$\displaystyle f'(4)=\lim_{h \to 0} \frac{f(4+h)-f(4)}{h}$

$\displaystyle =\lim_{h \to 0} \frac{\frac{3}{1+\sqrt{4+h}}-\frac{3}{1+\sqrt{4}}}{h}$

$\displaystyle =\lim_{h \to 0} \frac{\frac{3}{1+\sqrt{4+h}}-1}{h} \cdot \frac{1+\sqrt{4+h}}{1+\sqrt{4+h}}$

$\displaystyle =\lim_{h \to 0} \frac{3-(1+\sqrt{4+h})}{h(1+\sqrt{4+h})}$

$\displaystyle =\lim_{h \to 0} \frac{2-\sqrt{4+h}}{h(1+\sqrt{4+h})} \cdot \frac{2+\sqrt{4+h}}{2+\sqrt{4+h}}$

(it's multiplying by the conjugate of the numerator and then use the identity $\displaystyle (a-b)(a+b)=a^2-b^2)$

$\displaystyle =\lim_{h \to 0} \frac{4-(4+h)}{h(1+\sqrt{4+h})(2+\sqrt{4+h})}$

$\displaystyle =\lim_{h \to 0} \frac{-h}{h(1+\sqrt{4+h})(2+\sqrt{4+h})}$

$\displaystyle =\lim_{h \to 0} \frac{-1}{(1+\sqrt{4+h})(2+\sqrt{4+h})}$

And now the limit is defined...