If e^ θ =c+4 θ , where c is an arbitrary constant and is a function of θ , then what is d θ equal to?

Correct answer: B. -θ d aligned & e^ θ φ =C+4 θ · φ & ( θ · φ) e= (C+4 θ · φ) & θ · φ= (C+4 θ · φ) & θ · g φ=4\ (θ)+ (φ)\ + C aligned Differentiating both sides with respect to ' θ '. aligned & φ+θ · (d φ)/(d θ)=4((1)/(θ)+(1)/(φ) · (d φ)/(d θ)) & φ · d θ+θ · d φ=4 φ · d θ+4 · θ · d φ aligned aligned & 3 θ · d φ=-3 φ · d θ & φ · d θ=-θ · d φ aligned

NDA 2020

Exam: NDA
Year: 2020
Subject: Mathematics
Chapter: Differentiation

NDA 2020 · Mathematics · Differentiation

If e^ θ =c+4 θ , where c is an arbitrary constant and is a function...

e^{\theta \varphi}=c+4 \theta \varphi

, where

c

is an arbitrary constant and

\varphi

is a function of

\theta

, then what is

\varphi \mathrm{d} \theta

equal to?

Options

A $\theta \mathrm{d} \varphi$
B $-\theta \mathrm{d} \varphi$
C $4 \theta \mathrm{ d} \varphi$
D $-4 \theta \mathrm{ d} \varphi$

Correct answer

B. $-\theta \mathrm{d} \varphi$

Step-by-step solution

\begin{aligned} & e^{\theta \phi}=C+4 \theta \cdot \phi \\ & \text{ ( } \theta \cdot \phi) \ln e=\ln (C+4 \theta \cdot \phi) \\ & \theta \cdot \phi=\ln (C+4 \theta \cdot \phi) \\ & \Rightarrow \theta \cdot g \phi=4\{\ln (\theta)+\ln (\phi)\}+\ln C \end{aligned}

Differentiating both sides with respect to '

\theta

\begin{aligned} & \phi+\theta \cdot \frac{d \phi}{d \theta}=4\left(\frac{1}{\theta}+\frac{1}{\phi} \cdot \frac{d \phi}{d \theta}\right) \\ & \phi \cdot d \theta+\theta \cdot d \phi=4 \phi \cdot d \theta+4 \cdot \theta \cdot d \phi\end{aligned}

\begin{aligned} & \Rightarrow 3 \theta \cdot d \phi=-3 \phi \cdot d \theta \\ & \therefore \phi \cdot d \theta=-\theta \cdot d \phi\end{aligned}

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