## anonymous 5 years ago The graph of the derivative of a linear function is a parabola a straight line parallel to the x-axis a straight line parallel to the y-axis ............................

1. anonymous

hold on..... how to do dis type of quesn....... $x ^{x ^{}}$

2. dumbcow

linear function defined as y = mx+b _ > dy/dx = m m is constant and graph of constant is a straight horizontal line

3. anonymous

*...........$x ^{x ^{x}}$

4. anonymous

thnx dumbcow......

5. anonymous

$\text{Let } y = x^x$ $\implies \ln y = x \ln x$ And differentiate implicitly. Same applies to higher powers, just repeat.

6. anonymous

You could also write $x^x = e^{\ln x^x} = e^{x \ln x}$

7. anonymous

got it!!!!!!!!

8. anonymous

how to do dis 1........$x ^{^{x ^{x}}}$

9. anonymous

sorry but i m unable 2 understand

10. anonymous

$\text{Let } y = x^x$ $\implies \ln y = x \ln x \implies \frac{1}{y}\frac{\mathbb{d}y}{\mathbb{d}x} = \ln x + 1$ $\implies \frac{\mathbb{d}}{\mathbb{d}x} x^x = x^x(\ln x+1)$ $\text{Now let } z = x^{x^x}$ $\implies \ln z = x^x \ln x\ \implies \frac{1}{z}\frac{\mathbb{d}z}{\mathbb{d}x} = \frac{\mathbb{d}}{\mathbb{d}x} x^x \cdot \ln x + \frac{\mathbb{d}}{\mathbb{d}x} \ln x \cdot x^x$ But from above we know $\frac{\mathbb{d}}{\mathbb{d}x} x^x = x^x(\ln x+1)$ $\implies \frac{1}{z}\frac{\mathbb{d}z}{\mathbb{d}x} = x^x \left( \ln x \cdot (\ln x + 1) + \frac{1}{x} \right)$ $\implies \frac{\mathbb{d}}{\mathbb{d}x} x^{x^x} = x^{x^x}\left[ x^x \left( \ln x \cdot (\ln x + 1) + \frac{1}{x} \right) \right]$ Sorry for any typos.

11. anonymous

Similarly $\frac{\mathbb{d}}{\mathbb{d}x} x^{x^{x^{x}}} = x^{x^{x^{x}}} \left[ x^{x^x}\left[ \ln x \left( x^x \left( \ln x \cdot (\ln x + 1) + \frac{1}{x} \right)\right) + \frac{1}{x} \right] \right]$ It is left to the readers to generalise this for $x^{x^{x^{x^{x^{x^{x^{.^{.^{.}}}}}}}}}$