Find dy/dx y=arcsin(4x^5)
The problem you've presented is a calculus problem involving differentiation. Specifically, you are asked to determine the derivative of the function y with respect to the variable x, where y is defined as the inverse sine (arcsin) of a function of x, namely 4x^5. Essentially, you are being requested to compute the rate of change of y with respect to changes in x, using the rules of differentiation for composite functions, which in this case includes applying the chain rule and the derivative of the inverse sine function.
$y = arcsin \left(\right. 4 x^{5} \left.\right)$
Answer
Solution:
Step 1:
Take the derivative of both sides with respect to $x$: $\frac{d}{dx}(y) = \frac{d}{dx}(\arcsin(4x^5))$.
Step 2:
The derivative of $y$ with respect to $x$ is denoted as $\frac{dy}{dx}$.
Step 3:
Proceed to differentiate the right-hand side of the equation.
Step 3.1:
Utilize the chain rule, which in general form is $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$. Here, let $f(x) = \arcsin(x)$ and $g(x) = 4x^5$.
Step 3.1.1:
Introduce a substitution $u = 4x^5$ and differentiate: $\frac{d}{du}(\arcsin(u)) \cdot \frac{d}{dx}(4x^5)$.
Step 3.1.2:
The derivative of $\arcsin(u)$ with respect to $u$ is $\frac{1}{\sqrt{1-u^2}}$: $\frac{1}{\sqrt{1-u^2}} \cdot \frac{d}{dx}(4x^5)$.
Step 3.1.3:
Substitute $u$ back with $4x^5$: $\frac{1}{\sqrt{1-(4x^5)^2}} \cdot \frac{d}{dx}(4x^5)$.
Step 3.2:
Perform the differentiation.
Step 3.2.1:
Extract the constant $4$ from the derivative: $\frac{1}{\sqrt{1-(4x^5)^2}} \cdot 4 \cdot \frac{d}{dx}(x^5)$.
Step 3.2.2:
Simplify the expression.
Step 3.2.2.1:
Apply the product rule to $4x^5$: $\frac{1}{\sqrt{1-16(x^5)^2}} \cdot 4 \cdot \frac{d}{dx}(x^5)$.
Step 3.2.2.2:
Square the constant $4$: $\frac{1}{\sqrt{1-16x^{10}}} \cdot 4 \cdot \frac{d}{dx}(x^5)$.
Step 3.2.2.3:
Apply the power rule to $(x^5)^2$.
Step 3.2.2.3.1:
Use the power rule $(a^m)^n = a^{mn}$: $\frac{1}{\sqrt{1-16x^{10}}} \cdot 4 \cdot \frac{d}{dx}(x^5)$.
Step 3.2.2.3.2:
Multiply the exponents $5$ and $2$: $\frac{1}{\sqrt{1-16x^{10}}} \cdot 4 \cdot \frac{d}{dx}(x^5)$.
Step 3.2.3:
Since $4$ is a constant, the derivative of $4x^5$ is $4 \cdot \frac{d}{dx}(x^5)$: $\frac{1}{\sqrt{1-16x^{10}}} \cdot (4 \cdot \frac{d}{dx}(x^5))$.
Step 3.2.4:
Combine the constant $4$ with the fraction: $\frac{4}{\sqrt{1-16x^{10}}} \cdot \frac{d}{dx}(x^5)$.
Step 3.2.5:
Differentiate $x^5$ using the power rule: $\frac{4}{\sqrt{1-16x^{10}}} \cdot (5x^4)$.
Step 3.2.6:
Combine the terms.
Step 3.2.6.1:
Multiply $5$ with the fraction: $\frac{20}{\sqrt{1-16x^{10}}} \cdot x^4$.
Step 3.2.6.2:
Simplify the multiplication: $\frac{20x^4}{\sqrt{1-16x^{10}}}$.
Step 3.2.6.3:
Write the final expression: $\frac{20x^4}{\sqrt{1-16x^{10}}}$.
Step 4:
Express the derivative $\frac{dy}{dx}$ as equal to the right side: $\frac{dy}{dx} = \frac{20x^4}{\sqrt{1-16x^{10}}}$.
Step 5:
Substitute $\frac{dy}{dx}$ for $y$: $\frac{dy}{dx} = \frac{20x^4}{\sqrt{1-16x^{10}}}$.
Knowledge Notes:
To solve this problem, several calculus concepts are used:
Derivative of a Function: The derivative represents the rate at which a function is changing at any given point and is a fundamental concept in differential calculus.
Chain Rule: A technique for differentiating composite functions. If $f$ and $g$ are functions, then the derivative of $f(g(x))$ is $f'(g(x)) \cdot g'(x)$.
Derivative of Inverse Trigonometric Functions: The derivative of $\arcsin(x)$ is $\frac{1}{\sqrt{1-x^2}}$.
Power Rule: A basic rule for differentiation. If $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$.
Simplifying Expressions: Involves algebraic manipulation, such as factoring out constants and combining like terms, to make expressions easier to differentiate or integrate.
Substitution: A method where a part of the expression is replaced with a new variable to simplify the differentiation process.
By applying these concepts in the steps outlined above, the derivative of $y = \arcsin(4x^5)$ with respect to $x$ is found.
