Question

Find the slope of the tangent line for the curve, $f(x)= \dfrac{(2^x - 2^{-x})}{(2^x + 2^{-x})}$ at xy = 0.​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm \: f(x) = \dfrac{ {2}^{x} -{2}^{ - x} }{{2}^{x} + {2}^{ - x}}$

On differentiating both sides w. r. t. x, we get

$\rm \:\dfrac{d}{dx} f(x) =\dfrac{d}{dx} \: \dfrac{ {2}^{x} -{2}^{ - x} }{{2}^{x} + {2}^{ - x}}$

We know,

$\boxed{\tt{ \dfrac{d}{dx} \frac{u}{v} \: = \: \frac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2} } \: }}$

So, using this

$\rm \: f'(x) = \dfrac{({2}^{x} + {2}^{ - x})\dfrac{d}{dx}({2}^{x} - {2}^{ - x}) - ({2}^{x} - {2}^{ - x})\dfrac{d}{dx}({2}^{x} + {2}^{ - x})}{ {({2}^{x} + {2}^{ - x})}^{2} }$

We know,

$\boxed{\tt{ \: \dfrac{d}{dx} {a}^{x} \: = \: {a}^{x} \: loga \: }}$

So, using this result, we get

$\rm \: = \: \dfrac{({2}^{x} + {2}^{ - x})({2}^{x}log2 + {2}^{ - x}log2) - ({2}^{x} - {2}^{ - x})({2}^{x}log2 - {2}^{ - x}log2)}{ {({2}^{x} + {2}^{ - x})}^{2} }$

$\rm \: = \: \dfrac{log2({2}^{x} + {2}^{ - x})({2}^{x}+ {2}^{ - x}) - log2({2}^{x} - {2}^{ - x})({2}^{x} - {2}^{ - x})}{ {({2}^{x} + {2}^{ - x})}^{2} }$

$\rm \: = \: \dfrac{log2[ {({2}^{x} + {2}^{ - x})}^{2} - {({2}^{x} - {2}^{ - x})]}^{2} }{ {({2}^{x} + {2}^{ - x})}^{2} }$

We know,

$\boxed{\tt{ \: {(x + y)}^{2} - {(x - y)}^{2} = 4xy \: }}$

So, using this identity, we get

$\rm \: = \: \dfrac{log2(4 \times {2}^{x} \times {2}^{ - x})}{ {({2}^{x} + {2}^{ - x})}^{2} }$

$\rm \: = \: \dfrac{4 \: log2}{ {({2}^{x} + {2}^{ - x})}^{2} }$

We know,

If y = f(x) be any curve, then slope of tangent at point (a, b) on the curve is given bt f'(a).

So,

$\rm \: Slope \: of \: tangent, \: m \: = \: f'(0)$

$\rm \: = \: \dfrac{4 \: log2}{ {({2}^{0} + {2}^{0})}^{2} }$

$\rm \: = \: \dfrac{4 \: log2}{ {(1 + 1)}^{2} }$

$\rm \: = \: \dfrac{4 \: log2}{ {(2)}^{2} }$

$\rm \: = \: \dfrac{4 \: log2}{ 4 }$

$\rm \: = \: log2$

Hence,

$\rm\implies \:\rm \: Slope \: of \: tangent \: to \: f(x) = \dfrac{{2}^{x} - {2}^{ - x}}{{2}^{x} + {2}^{ - x}} \: is \: log2$

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ADDITIONAL INFORMATION

Let y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P (a, b) is called tangent and that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P. Slope of tangent is f'(a) and Slope of normal is - 1/f'(a)

2. If tangent is parallel to x - axis, its slope is 0.

3. If tangent is parallel to y - axis, its slope is not defined.

4. Two lines having slope M and m are parallel, iff M = m.

5. If two lines having slope M and m are perpendicular, iff Mm = - 1.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$