Find the derivative of cot^(-1)((1)/(8x+9)).

Find Inverse Cotangent Derivative: We need to find the derivative of the inverse cotangent function, which is cot^(-1)((1)/(8x+9)). Let's denote u = (1)/(8x+9). The derivative of cot^(-1)(u) with respect to u is -1/(1+u^2). We will use the chain rule to find the derivative with respect to x.Derivative of u: First, find the derivative of u = (1)/(8x+9) with respect to x. The derivative of 1/(8x+9) is -8/(8x+9)^2, using the power rule and the chain rule.Apply Chain Rule: Now, apply the chain rule. The derivative of cot^(-1)(u) with respect to x is the derivative of cot^(-1)(u) with respect to u multiplied by the derivative of u with respect to x. This gives us -1/(1+u^2) * (-8/(8x+9)^2).Substitute u: Substitute u back into the expression to get the derivative with respect to x. The derivative of cot^(-1)((1)/(8x+9)) with respect to x is -1/(1+(1/(8x+9))^2) * (-8/(8x+9)^2).Simplify Expression: Simplify the expression. The denominator of the first fraction becomes 1 + (1/(8x+9))^2 = (8x+9)^2/(8x+9)^2 + 1/(8x+9)^2 = ((8x+9)^2 + 1)/(8x+9)^2. The derivative simplifies to -1/(((8x+9)^2 + 1)/(8x+9)^2) * (-8/(8x+9)^2).Combine Fractions: Combine the fractions by multiplying the numerators and denominators. This gives us (-1 * -8)/(((8x+9)^2 + 1)/(8x+9)^2 * (8x+9)^2) which simplifies to 8/((8x+9)^2 + 1).

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

Find the derivative of $\cot^{-1}\left(\frac{1}{8x+9}\right)$ .

View step-by-step help

Home

Math Problems

Precalculus

Evaluate rational exponents

Full solution

Q. Find the derivative of

\cot^{-1}\left(\frac{1}{8x+9}\right)

Find Inverse Cotangent Derivative: We need to find the derivative of the inverse cotangent function, which is $\cot^{-1}\left(\frac{1}{8x+9}\right)$ . Let's denote $u = \frac{1}{8x+9}$ . The derivative of $\cot^{-1}(u)$ with respect to $u$ is $-\frac{1}{1+u^2}$ . We will use the chain rule to find the derivative with respect to $x$ .
Derivative of u: First, find the derivative of $u = \frac{1}{8x+9}$ with respect to $x$ . The derivative of $\frac{1}{8x+9}$ is $-\frac{8}{(8x+9)^2}$ , using the power rule and the chain rule.
Apply Chain Rule: Now, apply the chain rule. The derivative of $\cot^{-1}(u)$ with respect to $x$ is the derivative of $\cot^{-1}(u)$ with respect to $u$ multiplied by the derivative of $u$ with respect to $x$ . This gives us $-\frac{1}{1+u^2} \cdot \left(-\frac{8}{(8x+9)^2}\right)$ .
Substitute $u$ : Substitute $u$ back into the expression to get the derivative with respect to $x$ . The derivative of $\cot^{-1}\left(\frac{1}{8x+9}\right)$ with respect to $x$ is $-\frac{1}{1+\left(\frac{1}{8x+9}\right)^2} \cdot \left(-\frac{8}{(8x+9)^2}\right)$ .
Simplify Expression: Simplify the expression. The denominator of the first fraction becomes $1 + \left(\frac{1}{8x+9}\right)^2 = \frac{(8x+9)^2}{(8x+9)^2} + \frac{1}{(8x+9)^2} = \frac{((8x+9)^2 + 1)}{(8x+9)^2}$ . The derivative simplifies to $-\frac{1}{\left(\frac{(8x+9)^2 + 1}{(8x+9)^2}\right)} \times \left(-\frac{8}{(8x+9)^2}\right)$ .
Combine Fractions: Combine the fractions by multiplying the numerators and denominators. This gives us $(-1 \times -8)/(((8x+9)^2 + 1)/(8x+9)^2 \times (8x+9)^2)$ which simplifies to $8/((8x+9)^2 + 1)$ .