Let H ( X ) = U ( X ) V ( X ) H(x)=u(x)^{v(x)} H ( X ) = U ( X ) V ( X ) .(a) Using The Fact That Ln ⁡ [ U ( X ) V ( X ) ] = V ( X ) Ln ⁡ U ( X \ln \left[u(x)^{v(x)}\right]=v(x) \ln U(x Ln [ U ( X ) V ( X ) ] = V ( X ) Ln U ( X ], Use The Chain Rule, The Product Rule, And The Formula For The Derivative Of Ln ⁡ X \ln X Ln X To Show That$\frac{d}{d X} \ln

Mar 8, 2025 by ADMIN 356 views

$67. Let $h(x)=u(x)^{v(x)}$.$

(a) Derivative of $\ln h(x)$

To find the derivative of $\ln h(x)$ , we can use the chain rule, the product rule, and the formula for the derivative of $\ln x$ . First, let's recall the formula for the derivative of $\ln x$ :

\frac{d}{dx} \ln x = \frac{1}{x}

Now, let's use the chain rule to find the derivative of $\ln h(x)$ :

\frac{d}{dx} \ln h(x) = \frac{d}{dx} \ln \left[u(x)^{v(x)}\right]

Using the fact that $\ln \left[u(x)^{v(x)}\right]=v(x) \ln u(x)$ , we can rewrite the above equation as:

\frac{d}{dx} \ln h(x) = \frac{d}{dx} \left[v(x) \ln u(x)\right]

Now, let's use the product rule to find the derivative of $v(x) \ln u(x)$ :

\frac{d}{dx} \left[v(x) \ln u(x)\right] = v(x) \frac{d}{dx} \ln u(x) + \ln u(x) \frac{d}{dx} v(x)

Using the formula for the derivative of $\ln x$ , we can rewrite the above equation as:

\frac{d}{dx} \left[v(x) \ln u(x)\right] = v(x) \frac{1}{u(x)} \frac{d}{dx} u(x) + \ln u(x) \frac{d}{dx} v(x)

Now, let's use the chain rule to find the derivative of $v(x)$ :

\frac{d}{dx} v(x) = \frac{d}{dx} \left[v(x)\right]

Using the chain rule, we can rewrite the above equation as:

\frac{d}{dx} v(x) = \frac{d}{dx} \left[v(x)\right] = v'(x)

Now, let's substitute the above equation into the previous equation:

\frac{d}{dx} \left[v(x) \ln u(x)\right] = v(x) \frac{1}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v'(x)

Now, let's use the product rule to find the derivative of $v(x) \ln u(x)$ :

\frac{d}{dx} \left[v(x) \ln u(x)\right] = v(x) \frac{1}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v'(x)

Now, let's substitute the above equation into the previous equation:

\frac{d}{dx} \ln h(x) = v(x) \frac{1}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v'(x)

(b) Simplifying the Derivative

Now, let's simplify the derivative of $\ln h(x)$ :

\frac{d}{dx} \ln h(x) = v(x) \frac{1}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v'(x)

Using the fact that $h(x)=u(x)^{v(x)}$ , we can rewrite the above equation as:

\frac{d}{dx} \ln h(x) = v(x) \frac{1}{u(x)} \frac{d}{dx} u(x) + \ln u(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Now, let's use the chain rule to find the derivative of $u(x)^{v(x)}$ :

\frac{d}{dx} \left[u(x)^{v(x)}\right] = \frac{d}{dx} \left[u(x)^{v(x)}\right] = v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Using the chain rule, we can rewrite the above equation as:

\frac{d}{dx} \left[u(x)^{v(x)}\right] = v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right] = v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Now, let's substitute the above equation into the previous equation:

\frac{d}{dx} \ln h(x) = v(x) \frac{1}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Now, let's simplify the derivative of $\ln h(x)$ :

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

(c) Final Answer

Now, let's simplify the derivative of $\ln h(x)$ :

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Using the fact that $h(x)=u(x)^{v(x)}$ , we can rewrite the above equation as:

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} h(x)

Now, let's simplify the derivative of $\ln h(x)$ :

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} h(x)

Using the fact that $h(x)=u(x)^{v(x)}$ , we can rewrite the above equation as:

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Now, let's simplify the derivative of $\ln h(x)$ :

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Using the chain rule, we can rewrite the above equation as:

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Now, let's simplify the derivative of $\ln h(x)$ :

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Using the fact that $h(x)=u(x)^{v(x)}$ , we can rewrite the above equation as:

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} h(x)

Now, let's simplify the derivative of $\ln h(x)$ :

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} h(x)

Using the fact that $h(x)=u(x)^{v(x)}$ , we can rewrite the above equation as:

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Now, let's simplify the derivative of $\ln h(x)$ :

\frac{d}{dx} \ln h(x) = \frac{v(x)}{u(x)} \frac{d}{dx} u(x) + \ln u(x) v(x) \frac{d}{dx} \left[u(x)^{v(x)}\right]

Using the chain rule, we can rewrite the above equation as:

\frac{d}{<br/> # 67. Let $h(x)=u(x)^{v(x)}$. Q&A

(a) Q: What is the derivative of $\ln h(x)$ ?

$(a) Q: What is the derivative of $\ln h(x)$?$

A: To find the derivative of $\ln h(x)$ , we can use the chain rule, the product rule, and the formula for the derivative of $\ln x$ . First, let's recall the formula for the derivative of $\ln x$ :

(a) Derivative of ln⁡h(x)\ln h(x)lnh(x)

(b) Simplifying the Derivative

(c) Final Answer

(a) Q: What is the derivative of ln⁡h(x)\ln h(x)lnh(x)?

(a) Derivative of $\ln h(x)$

(a) Q: What is the derivative of $\ln h(x)$ ?