Derivative of y w.r.t x: 242.7x.25

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In summary, the derivative of $\displaystyle y=8\ln{x}+\sqrt{1-x^2}\arccos{x}$ with respect to $x$ is $\displaystyle \frac{8}{x}-\frac{\sqrt{1-x^2}}{\sqrt{1-x^2}}=\frac{8}{x}-1$.
  • #1
karush
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$\tiny{242.7x.25}$
$\textsf{Find the derivative of y with respect to x}$
\begin{align*}\displaystyle
y&=8\ln{x}+\sqrt{1-x^2}\arccos{x} \\
&=\frac{8}{x}+?
\end{align*}
 
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  • #2
We could use a table of forumlas here, but let's derive the formula we need. Let:

\(\displaystyle y=\arccos(x)\)

\(\displaystyle \cos(y)=x\)

Implicitly differentiate:

\(\displaystyle -\sin(y)\d{y}{x}=1\)

\(\displaystyle \d{y}{x}=-\csc(y)=-\csc(\arccos(x))=-\frac{1}{\sqrt{1-x^2}}\)

Can you proceed with the product rule?
 
  • #3
MarkFL said:
We could use a table of forumlas here, but let's derive the formula we need. Let:

\(\displaystyle y=\arccos(x)\)

\(\displaystyle \cos(y)=x\)

Implicitly differentiate:

\(\displaystyle -\sin(y)\d{y}{x}=1\)

\(\displaystyle \d{y}{x}=-\csc(y)=-\csc(\arccos(x))=-\frac{1}{\sqrt{1-x^2}}\)

Can you proceed with the product rule?

As it's not entirely obvious why $\displaystyle \begin{align*} \csc{ \left[ \arccos{ \left( x \right) } \right] } = \frac{1}{\sqrt{1 - x^2}} \end{align*}$

$\displaystyle \begin{align*} \csc{ \left[ \arccos{(x)} \right] } &= \frac{1}{\sin{\left[ \arccos{ (x) } \right] }} \\ &= \frac{1}{\sqrt{ 1 - \left\{ \cos{ \left[ \arccos{(x)} \right] } \right\} ^2 }} \\ &= \frac{1}{\sqrt{1 - x^2}} \end{align*}$
 

Related to Derivative of y w.r.t x: 242.7x.25

1. What is the meaning of "Derivative of y w.r.t x: 242.7x.25"?

The expression "Derivative of y w.r.t x: 242.7x.25" is a mathematical notation that represents the derivative of the function y with respect to the variable x. The value 242.7x.25 is a constant multiplier that affects the rate of change of the function y.

2. How do you calculate the derivative of y with respect to x?

The derivative of y with respect to x is calculated using the rules of differentiation, which involve taking the limit of the slope of a secant line as the distance between two points approaches zero. In this case, the constant multiplier 242.7x.25 is treated like any other constant when taking the derivative.

3. What is the significance of the constant multiplier in the expression?

The constant multiplier, 242.7x.25, affects the rate of change of the function y with respect to the variable x. It determines the slope of the tangent line at any given point on the function's graph, indicating how quickly the function is changing at that specific point.

4. Can you provide an example of how to calculate the derivative of y w.r.t x for a specific value of x?

Yes, for example, let's say y = 2x^2. To calculate the derivative of y with respect to x, we use the power rule for differentiation, which states that the derivative of x^n is nx^(n-1). In this case, the derivative of y with respect to x is 4x. So, for a specific value of x, such as x = 3, the derivative of y w.r.t x would be 4(3) = 12.

5. How is the derivative of y w.r.t x used in real-world applications?

The derivative of y w.r.t x has many practical applications in fields such as physics, engineering, economics, and finance. It can be used to calculate rates of change, find maximum and minimum values, and solve optimization problems. For example, in physics, the derivative of position with respect to time is velocity, and the derivative of velocity with respect to time is acceleration.

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