Find the directional derivative of q = 4x ^ 3 - 3x ^ 2 * y ^ 2 (2.-1.1) along the line which makes equal angles with Co-ordinate axes.

q =  4x³ - 3x²y²

\(\vec ∇q\) = (\(\hat i \frac{\delta}{\delta x}\) + \(\hat j \frac{\delta}{\delta y}\) + \(\hat k \frac{\delta}{\delta z}\))q

= \(\hat i \frac{\delta}{\delta x}\)(4x³ - 3x²y²) + \(\frac{j\delta}{\delta y}\)(4x³ - 3x²y²) + \(0\hat k\)

= \(\hat i\) (12x² - 6xy²) + \(\hat j\)(-6x²y)

Direction cosines of line which makes equal angles with coordinate axis are (±\(\frac{1}{\sqrt 3}\), ±\(\frac{1}{\sqrt 3}\),±\(\frac{1}{\sqrt 3}\)).

∴ Direction derivative of q in direction of line is

= (\(\vec ∇q\)).(\(\frac{1}{\sqrt3}\hat i\) + \(\frac{1}{\sqrt3}\hat j\) + \(\frac{1}{\sqrt3}\hat k\))

= \(\hat i\) (\(12x^2-6xy^2\)) + \(\hat j\) (-6x²y)(\(\frac{1}{\sqrt3}\hat i\) + \(\frac{1}{\sqrt3}\hat j\) + \(\frac{1}{\sqrt3}\hat k\))

= \(\frac{1}{\sqrt 3}\)\((12x^2-6xy^2)\) - \(\frac{6x^2y}{\sqrt 3}\) 

Direction derivative at (2,-1,1) = \(\frac{1}{\sqrt 3}\)(12 x 4 - 6 x 2 x 1) - \(\frac{6\times 2^2\times -1}{\sqrt 3}\) 

= \(\frac{1}{\sqrt 3}\)(48 - 12) + \(\frac{24}{\sqrt 3}\) 

= \(\frac{60}{\sqrt 3}\) 

= \(20\sqrt 3\).