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anonymous
 5 years ago
The graph of the derivative of a linear function is
a parabola
a straight line parallel to the xaxis
a straight line parallel to the yaxis ............................
anonymous
 5 years ago
The graph of the derivative of a linear function is a parabola a straight line parallel to the xaxis a straight line parallel to the yaxis ............................

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hold on..... how to do dis type of quesn....... \[x ^{x ^{}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0linear function defined as y = mx+b _ > dy/dx = m m is constant and graph of constant is a straight horizontal line

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0*...........\[x ^{x ^{x}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\text{Let } y = x^x \] \[\implies \ln y = x \ln x \] And differentiate implicitly. Same applies to higher powers, just repeat.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0You could also write \[x^x = e^{\ln x^x} = e^{x \ln x} \]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how to do dis 1........\[x ^{^{x ^{x}}}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry but i m unable 2 understand

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\text{Let } y = x^x \] \[\implies \ln y = x \ln x \implies \frac{1}{y}\frac{\mathbb{d}y}{\mathbb{d}x} = \ln x + 1 \] \[\implies \frac{\mathbb{d}}{\mathbb{d}x} x^x = x^x(\ln x+1) \] \[\text{Now let } z = x^{x^x} \] \[\implies \ln z = x^x \ln x\ \implies \frac{1}{z}\frac{\mathbb{d}z}{\mathbb{d}x} = \frac{\mathbb{d}}{\mathbb{d}x} x^x \cdot \ln x + \frac{\mathbb{d}}{\mathbb{d}x} \ln x \cdot x^x \] 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.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Similarly \[ \frac{\mathbb{d}}{\mathbb{d}x} x^{x^{x^{x}}} = x^{x^{x^{x}}} \left[ x^{x^x}\left[ \ln x \left( x^x \left( \ln x \cdot (\ln x + 1) + \frac{1}{x} \right)\right) + \frac{1}{x} \right] \right] \] It is left to the readers to generalise this for \[x^{x^{x^{x^{x^{x^{x^{.^{.^{.}}}}}}}}} \]
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