Question

53-56 Find \(y'\) and \(y''\).

54. \(y = {\left( {{\bf{1}} + \sqrt x } \right)^{\bf{3}}}\)

Short answer

The value of \(y'\) is \(\frac{{3{{\left( {1 + \sqrt x } \right)}^2}}}{{2\sqrt x }}\).

The value of \(y''\) is \(\frac{{3\left( {x - 1} \right)}}{{4{x^{\frac{3}{2}}}}}\).

Step-by-step solution

Step 1:Find the value of \(y'\)

Find the value of \(y'\) for the equation\(y = {\left( {1 + \sqrt x } \right)^3}\).

\(\begin{aligned}\frac{{{\rm{d}}y}}{{{\rm{d}}x}} &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {{{\left( {1 + \sqrt x } \right)}^3}} \right)\\ &= 3{\left( {1 + \sqrt x } \right)^2} \cdot \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {1 + \sqrt x } \right)\\ &= \frac{{3{{\left( {1 + \sqrt x } \right)}^2}}}{{2\sqrt x }}\end{aligned}\)

So, the value of \(y'\) is \(\frac{{3{{\left( {1 + \sqrt x } \right)}^2}}}{{2\sqrt x }}\).

Find the value of \(y''\)

Find the value of \(y''\) from the equation \(y' = \frac{{3{{\left( {1 + \sqrt x } \right)}^2}}}{{2\sqrt x }}\) using the quotient rule.

\(\begin{aligned}y'' &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{3{{\left( {1 + \sqrt x } \right)}^2}}}{{2\sqrt x }}} \right)\\ &= \frac{{2\sqrt x \cdot \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {3{{\left( {1 + \sqrt x } \right)}^2}} \right) - 3{{\left( {1 + \sqrt x } \right)}^2} \cdot \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {2\sqrt x } \right)}}{{{{\left( {2\sqrt x } \right)}^2}}}\\ &= \frac{{2\sqrt x \left( {6\left( {1 + \sqrt x } \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {1 + \sqrt x } \right)} \right) - 3{{\left( {1 + \sqrt x } \right)}^2} \times \frac{2}{{2\sqrt x }}}}{{4x}}\\ &= \frac{{12\sqrt x \left( {1 + \sqrt x } \right) \times \frac{1}{{2\sqrt x }} - \frac{{3{{\left( {1 + \sqrt x } \right)}^2}}}{{\sqrt x }}}}{{4x}}\\ &= \frac{{6 + 6\sqrt x - \frac{{3\left( {1 + x + 2\sqrt x } \right)}}{{\sqrt x }}}}{{4x}}\\ &= \frac{{6\sqrt x + 6x - 3 - 3x - 6\sqrt x }}{{4{x^{\frac{3}{2}}}}}\\ &= \frac{{3\left( {x - 1} \right)}}{{4{x^{\frac{3}{2}}}}}\end{aligned}\)

So, the value of \(y''\) is \(\frac{{3\left( {x - 1} \right)}}{{4{x^{\frac{3}{2}}}}}\).