Question

Differentiate y = x arcsin(5x) with respect to x

Answer 1

The derivative of y = x arcsin(5x) with respect to x is calculated using the product rule, resulting in the derivative being arcsin(5x) + 5x/sqrt(1-(5x)^2).

Step-by-step explanation:

To differentiate the function y = x arcsin(5x) with respect to x, we will use the product rule of differentiation, which states that if u(x) and v(x) are functions of x, then the derivative of their product u(x)v(x) is u'(x)v(x) + u(x)v'(x). In our case, u(x) = x and v(x) = arcsin(5x). The derivative of u(x) is simply 1, and to differentiate v(x), we recall that the derivative of arcsin(u) is 1/sqrt(1-u^2) by the chain rule, and hence the derivative of v(x) is 5/sqrt(1-(5x)^2). Applying the product rule, we obtain:

`y' = \arcsin(5x) + x \cdot (5)/(√(1 - (5x)^2))`

Thus, the derivative of y with respect to x is
`\arcsin(5x) + (5x)/(√(1 - (5x)^2))`

Answer 2

The derivative of y = x arcsin(5x) with respect to x is:

y' = 5√(1 - 25x²) + x

Step-by-step explanation:

To differentiate y = x arcsin(5x), we need to employ the chain rule and the derivative of the inverse sine function (arcsin).

Chain Rule: Since y is a composite function involving x and arcsin(5x), we need to use the chain rule. The chain rule states:

d/dx[w(u(x))] = w'(u(x)) * u'(x)

where w and u are any differentiable functions.

In this case, y = w(u(x)) = x * u(x), where u(x) = arcsin(5x).

Differentiate u(x): The derivative of arcsin(x) is 1/√(1 - x²). However, we need to substitute 5x for x due to the composite function:

u'(x) = d/dx[arcsin(5x)] = 1/√(1 - (5x)²)

Differentiate w(u(x)): The derivative of w(u(x)) is simply u(x) itself:

w'(u(x)) = d/dx[x * u(x)] = u(x) = x

Apply the chain rule: Putting it all together using the chain rule:

y' = d/dx[x arcsin(5x)] = w'(u(x)) * u'(x)

= x * 1/√(1 - (5x)²)

= 5√(1 - 25x²) + x

Therefore, the derivative of y = x arcsin(5x) with respect to x is 5√(1 - 25x²) + x.