/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 6 $\begin{aligned} y &=\ln \le... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 6

$\begin{aligned} y &=\ln \left|\operatorname{sece}^{x^{\prime}}\right| \\\ y^{\prime} &=\frac{1}{\sec e^{x^{2}}} \times \sec e^{x^{3}} \tan e^{x^{2}} \times e^{x^{2}} \times 2 x \\ &=2 x e^{x^{2}} \tan e^{x^{3}} \end{aligned}$

Short Answer

Expert verified
Question: Find the derivative of the function \(y = \ln \left|\operatorname{sece}^{x^{\prime}}\right|\). Answer: The derivative of the given function is \(y^{\prime} = 2xe^{x^2}\tan{e^{x^2}}\).

Step by step solution

01

Understand the given function

We are given a function \(y = \ln \left|\operatorname{sece}^{x^{\prime}}\right|\). Our first goal is to differentiate this function with respect to x.
02

Apply the chain rule on the given function

To differentiate the natural logarithm of the absolute value function, we need to differentiate the natural logarithm of a function as follows: \(\frac{d}{dx}[\ln f(x)] = \frac{f^{\prime}(x)}{f(x)}\) Applying the chain rule on the given function, we have: \(y^{\prime} = \frac{d}{dx}[\ln \left|\operatorname{sece}^{x^{\prime}}\right|] = \frac{1}{\operatorname{sece}^{x^2}} \times \frac{d}{dx}[\operatorname{sece}^{x^2}]\)
03

Differentiate the secant function

We need to differentiate the exponential function with the secant function as its base: \(\frac{d}{dx}[\operatorname{sece}^{x^{2}}] = \sec(e^{x^2}) \tan(e^{x^2}) \times \frac{d}{dx}[e^{x^2}]\) Now we need to differentiate the exponential function inside the secant function: \(\frac{d}{dx}[e^{x^2}] = e^{x^2} \times \frac{d}{dx}[x^2] = e^{x^2} \times 2x\)
04

Combine all the results

Combining all the results from steps 2 and 3: \(y^{\prime} = \frac{1}{\operatorname{sece}^{x^2}} \times [\sec(e^{x^3}) \tan(e^{x^2}) \times (e^{x^2} \times 2x)]\) After simplifying this expression, we get: \(y^{\prime} = 2xe^{x^2}\tan e^{x^3}\)

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Most popular questions from this chapter

$\begin{aligned} &\text { A) } \begin{array}{l} y^{2}+4=\left(\sec ^{n} \theta+\cos ^{n} \theta\right)^{2} \quad \ \\ x^{2}+4=(\sec \theta+\cos \theta)^{2} \\ \frac{d x}{d \theta}=\sec \theta \tan \theta+\sin \theta \\ \frac{d y}{d \theta}=n \sec ^{n-1} \theta \sec \theta \tan \theta+n \cos ^{n-1} \theta \sin \theta \\ \left(\frac{d y}{d x}\right)^{2}=\frac{n^{2}\left(\sec ^{n} \theta \tan \theta+\cos ^{n} \theta \tan \theta\right)^{2}}{\tan ^{2} \theta(\sec \theta+\cos \theta)^{2}} \\ =\frac{n^{2}\left(y^{2}+4\right)}{x^{2}+4} \end{array} \end{aligned}$ $\begin{aligned} &\text { B) }\\\ &\text { Put } t=\tan \theta\\\ &x=0, \quad y=0\\\ &\frac{\mathrm{d} \mathrm{x}}{\mathrm{d} \theta}=1 \quad \frac{\mathrm{dy}}{\mathrm{d} \theta}=1\\\ &\Rightarrow \frac{d y}{d x}=1 \end{aligned}$ $\begin{aligned} &\text { C) }\\\ &\mathrm{e}^{y}+x y=e\\\ &\begin{aligned} &\mathrm{e}^{y} \mathrm{y}^{\prime}+\mathrm{xy}^{\prime}+\mathrm{y}=0 \\ &\mathrm{e}^{\mathrm{y}} \mathrm{y}^{\prime \prime}+\mathrm{e}^{y}\left(\mathrm{y}^{\prime}\right)^{2}+\mathrm{y}^{\prime}+\mathrm{xy}^{\prime \prime}+\mathrm{y}^{\prime}=0 \end{aligned} \end{aligned}$ $\begin{aligned} &\text { For } x=0, y=1 \\ &y^{\prime \prime}=\frac{-\left(e\left(y^{\prime}\right)^{2}+2 y^{\prime}\right)}{e} \\ &=\frac{-1}{e}\left(\frac{1}{e}-\frac{2}{e}\right)=\frac{1}{e^{2}} \end{aligned}$ $\begin{aligned} &\text { D) } \phi(x)=f(x) g(x) \\ &\phi^{\prime}(x)=f^{\prime}(x) g(x)+f(x) g^{\prime}(x) \\ &\phi^{\prime \prime}(x)=f^{\prime \prime}(x) g(x)+2 f^{\prime}(x) g^{\prime}(x)+f(x) g^{\prime}(x) \\ &\frac{\phi^{\prime \prime}(x)}{\phi(x)}=\frac{f^{\prime \prime}(x)}{f(x)}+\frac{g^{\prime \prime}(x)}{g(x)}+\frac{2 f^{\prime}(x)}{f(x)} \frac{g^{\prime}(x)}{g(x)} \end{aligned}$

As $\mathrm{f}(\mathrm{x}), \mathrm{f}^{\prime}(\mathrm{x}), \mathrm{f}^{\prime \prime}(\mathrm{x})\( are all \)+\mathrm{ve} \forall \mathrm{x} \in[0,7]$ \(\Rightarrow \mathrm{f}(\mathrm{x})\) is increasing function and concave up. and so is \(\mathrm{f}^{-1}(\mathrm{x})\). $\Rightarrow f^{-1}(5)+4 f^{-1}\left(\frac{2 g}{5}\right)\( is always \)+v e$.

\((f(x))^{n}=f(n x)\) Differentiating it \((f(x))^{n-1} f^{\prime}(x)=f^{\prime}(n x)\) Multiply by \(f(x)\) $\mathrm{f}(\mathrm{nx}) \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}^{\prime}(\mathrm{nx}) \mathrm{f}(\mathrm{x})$

$\begin{aligned} &\frac{d^{2} x}{d y^{2}}\left(\frac{d y}{d x}\right)^{3}+\frac{d^{2} y}{d x^{2}}=k \\ &\text { As } \frac{d x}{d y}=\frac{1}{d y / d x} \\ &\Rightarrow \frac{d^{2} x}{d y^{2}}=\frac{-d^{2} y / d x^{2}}{(d y / d x)^{3}} \\ &\Rightarrow \frac{d^{2} x}{d y^{2}}\left(\frac{d y}{d x}\right)^{3}+\frac{d^{2} y}{d x^{2}}=0 \end{aligned}$

Column-I (A) If \(|g(x)-g(y)| \leq|x-y|^{99},(n \in N)\) then \(g^{\prime}(x)\) equals (B) Differential coefficient of $\sin ^{-1} \frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\( with respect to \)\cos ^{-1} \frac{1-x^{2}}{1+x^{2}}\( in the domain of \)f(x)=\frac{1}{\sqrt{1-x^{2}}}$ is (C) If \(y=\cot ^{-1}(\tan x)+\tan ^{-1}(\cot x), \pi / 2
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