Question

Find equations of the tangent line to the given curve at the specific point.

35. \(y = \frac{{{x^2}}}{{1 + x}}\), \(\left( {1,\frac{1}{2}} \right)\)

Short answer

The equation of the tangent line is \(y = \frac{3}{4}x - \frac{1}{4}\).

Step-by-step solution

The slope of the tangent

The slope of the tangent line to the curve is equal to the derivative of the function representing that curve at some given point.

The quotient rule of differentiation

If a function \(h\left( x \right) = \frac{{f\left( x \right)}}{{g\left( x \right)}}\) where \(f\) and \(g\) are both differentiable, then the derivative of \(h\left( x \right)\) is as follows:

\(\begin{aligned}\frac{{dh}}{{dx}} & = \frac{d}{{dx}}\left( {\frac{{f\left( x \right)}}{{g\left( x \right)}}} \right)\\ & = \frac{{g\left( x \right) \cdot \frac{d}{{dx}}\left( {f\left( x \right)} \right) - f\left( x \right) \cdot \frac{d}{{dx}}\left( {g\left( x \right)} \right)}}{{{{\left( {g\left( x \right)} \right)}^2}}}\end{aligned}\)

The derivative of the given function

Differentiate \(f\left( x \right) = \frac{{{x^2}}}{{1 + x}}\) using the quotient rule as follows:

\(\begin{aligned}f\left( x \right) & = \frac{{{x^2}}}{{1 + x}}\\f'\left( x \right) & = \frac{{\left( {1 + x} \right)\frac{d}{{dx}}\left( {{x^2}} \right) - {x^2}\frac{d}{{dx}}\left( {1 + x} \right)}}{{{{\left( {1 + x} \right)}^2}}}\\ & = \frac{{\left( {1 + x} \right)\left( {2x} \right) - {x^2}\left( 1 \right)}}{{{{\left( {1 + x} \right)}^2}}}\\ & = \frac{{2x + 2{x^2} - {x^2}}}{{{{\left( {1 + x} \right)}^2}}}\\ & = \frac{{2x + {x^2}}}{{{{\left( {1 + x} \right)}^2}}}\end{aligned}\)

Hence, the value of \(f'\left( x \right)\) is\(\frac{{2x + {x^2}}}{{{{\left( {1 + x} \right)}^2}}}\).

The slope of the tangent at \(\left( {1,\frac{1}{2}} \right)\)

Put \(x = 1\) in \(f'\left( x \right)\) and solve as follows:

\(\begin{aligned}f'\left( 1 \right) & = \frac{{2\left( 1 \right) + {{\left( 1 \right)}^2}}}{{{{\left( {1 + 1} \right)}^2}}}\\ & = \frac{{2 + 1}}{{{{\left( 2 \right)}^2}}}\\ & = \frac{3}{4}\end{aligned}\)

The equation of a line

The general equation of a line that passes through the point\(\left( {{x_1},{y_1}} \right)\)and has slope\(m\)is given below:

\(y - {y_1} = m\left( {x - {x_1}} \right)\)

The equation of a tangent line

Substitute the value of\({x_1}\),\({y_2}\)in the above formula to obtain the equation of tangent as follows:

\(\begin{aligned}y - \frac{1}{2} & = \frac{3}{4}\left( {x - 1} \right)\\y - \frac{1}{2} & = \frac{3}{4}x - \frac{3}{4}\\y & = \frac{3}{4}x - \frac{3}{4} + \frac{1}{2}\\y & = \frac{3}{4}x - \frac{1}{4}\end{aligned}\)

Hence, the equation of the tangent line is \(y = \frac{3}{4}x - \frac{1}{4}\).