Find the x-coordinate of the point at which the tangent to the curvey =x> 0is horizontal. Give the exact value of x. (Note that by the order of operations, y =not (x")5 = 25 ,ava|a|sin (a)T

Name: Find the x-coordinate of the point at which the tangent to the curvey
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Description: Find the x-coordinate of the point at which the tangent to the curvey =x> 0is horizontal. Give the exact value of x. (Note that by the order of operations, y =not (x")5 = 25 ,ava|a|sin (a)T

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Mathematics For Machine Technology

8th Edition

ISBN:9781337798310

Author:Peterson, John.

Publisher:Peterson, John.

Chapter81: Introduction To Computer Numerical Control (cnc)

Section: Chapter Questions

Problem 24A

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Find the x-coordinate of the point at which the tangent to the curve
y =
x> 0
is horizontal. Give the exact value of x. (Note that by the order of operations, y =
not (x")5 = 25 ,
a
va
|a|
sin (a)
T

Expert Solution

Step 1

Here, we need to find the x-coordinate of the point at which the tangent to the curve $y = x^{x^{5}}, x > 0$ is horizontal.

$We know the exponent rule : a^{b} = e^{b \ln (a)} .$

Now,

$\frac{d y}{d x} = \frac{d}{d x} (x^{x^{5}}) = \frac{d}{d x} (e^{x^{5} \ln (x)}) = e^{x^{5} \ln (x)} (x^{5} \ln (x))' (by the chain rule) = e^{x^{5} \ln (x)} (5 x^{4} \ln (x) + x^{4}) (by using the product rule) = x^{x^{5}} (5 x^{4} \ln (x) + x^{4}) (as e^{x^{5} \ln (x)} = x^{x^{5}})$

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