Question

Find \(D_{x} y\). $$ y=x^{2} \sinh ^{-1}\left(x^{5}\right) $$

Short answer

\( y' = 2x \sinh^{-1}(x^5) + \frac{5x^6}{\sqrt{x^{10} + 1}} \).

Step-by-step solution

Identify the Functions Involved

The function given is a product of two functions: \( y = u(x) v(x) \) where \( u(x) = x^2 \) and \( v(x) = \sinh^{-1}(x^5) \).

Differentiate Both Functions

Find the derivatives of the individual functions.1. \( u(x) = x^2 \) - Using the power rule, \( u'(x) = 2x \).2. \( v(x) = \sinh^{-1}(x^5) \) - Use the chain rule for differentiation. First, recall that \( \frac{d}{dx} \sinh^{-1}(x) = \frac{1}{\sqrt{x^2 + 1}} \). Therefore, \( v'(x) = \frac{d}{dx}(x^5) \times \frac{1}{\sqrt{(x^5)^2 + 1}} = 5x^4 \times \frac{1}{\sqrt{x^{10} + 1}} \).

Apply the Product Rule

The derivative of a product \( y = u(x)v(x) \) is given by \( y' = u'(x)v(x) + u(x)v'(x) \). Substitute \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \) into this formula.

Substitute and Simplify

Substitute the derivatives from Step 2 into the product rule from Step 3:\( y' = 2x \times \sinh^{-1}(x^5) + x^2 \times 5x^4 \times \frac{1}{\sqrt{x^{10} + 1}} \).Simplify the expression: \( y' = 2x \sinh^{-1}(x^5) + \frac{5x^6}{\sqrt{x^{10} + 1}} \).

Key concepts

Inverse hyperbolic functions

Inverse hyperbolic functions are analogous to inverse trigonometric functions, but they are based on hyperbolic functions. The key inverse hyperbolic functions are: inverse hyperbolic sine (\( \sinh^{-1}(x) = \ln(x + \sqrt{x^2 + 1}) \)), inverse hyperbolic cosine (\( \cosh^{-1}(x) = \ln(x + \sqrt{x^2 - 1}) \)), and inverse hyperbolic tangent (\( \tanh^{-1}(x) = \frac{1}{2} \ln(\frac{1 + x}{1 - x}) \)).
For this exercise, we focus on inverse hyperbolic sine, noted as \( \sinh^{-1}(x) \). It's important to remember the derivative of \( \sinh^{-1}(x) \):

• \( \frac{d}{dx} \sinh^{-1}(x) = \frac{1}{\sqrt{x^2 + 1}} \)

If you are working with a composite function, like \( \sinh^{-1}(x^5) \), you'll need to apply the chain rule to find its derivative. This is because the inner function \( x^5 \) affects the complexity of the derivative calculation.

Product rule

The product rule is used when differentiating two functions that are multiplied together. If you have two functions \( u(x) \) and \( v(x) \), their derivative, denoted as \( (uv)' \), can be calculated using:

• \( (uv)' = u'v + uv' \)

In this exercise, we identified \( u(x) = x^2 \) and \( v(x) = \sinh^{-1}(x^5) \). Using the product rule, the derivative of \( y = x^2 \sinh^{-1}(x^5) \) involves a combination of the derivatives \( u'(x) \) and \( v'(x) \), along with the original functions. So, the differential operation becomes \( y' = u'v + uv' \), which will be \( 2x \cdot \sinh^{-1}(x^5) + x^2 \cdot v'(x) \). Always remember to keep the order in check to ensure the solution is simplified correctly.

Chain rule

The chain rule is fundamental when dealing with composite functions. It helps distinguish when a function is embedded within another. Given a composite function \( f(g(x)) \), the derivative is found by:

• \( f'(g(x)) \times g'(x) \)

For this problem, we applied the chain rule to differentiate \( \sinh^{-1}(x^5) \). First, we know \( \frac{d}{dx} \sinh^{-1}(x) = \frac{1}{\sqrt{x^2 + 1}} \). Therefore, for \( \sinh^{-1}(x^5) \), the outer function's derivative with respect to its inner function \( x^5 \) is \( \frac{1}{\sqrt{(x^5)^2 + 1}} \), and the inner derivative of \( x^5 \) is \( 5x^4 \). Combining these gives the chain rule result: \( 5x^4 \cdot \frac{1}{\sqrt{x^{10} + 1}} \). This shows how derivatives of composite functions entail using chain rule efficiently.

Power rule

The power rule is a straightforward technique for differentiating functions of the form \( x^n \). For \( n \) being any real number, the derivative is:

• \( \frac{d}{dx} x^n = nx^{n-1} \)

In this exercise, applying the power rule was necessary to find the derivative of \( u(x) = x^2 \). We simply apply the derivative formula: \( 2x^{2-1} = 2x \). Power rules also apply wherever you see a pure power function. In our solution, it's crucial for obtaining parts of the derivative's components, particularly for extracting \( u'(x) \). It’s a foundational tool in calculus and helps simplify the process when coupled with other rules like the product and chain rules.