Find dy/dx y=arcsin(4x^5)

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 $\frac{d}{dx}[f(g(x))]=f^{\prime }(g(x))\cdot g^{\prime }(x)$. 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:

1. 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.

2. Chain Rule: A technique for differentiating composite functions. If $f$ and $g$ are functions, then the derivative of $f(g(x))$ is $f^{\prime }(g(x))\cdot g^{\prime }(x)$.

3. Derivative of Inverse Trigonometric Functions: The derivative of $\arcsin (x)$ is $\frac{1}{\sqrt{1-x^2}}$.

4. Power Rule: A basic rule for differentiation. If $n$ is a real number and $f(x)=x^n$, then $f^{\prime }(x)=n{x}^{n-1}$.

5. Simplifying Expressions: Involves algebraic manipulation, such as factoring out constants and combining like terms, to make expressions easier to differentiate or integrate.

6. 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.