Question

Find the derivative of Y with respect to theta.
Y=ln(theta -4)-e^theta

Answer 1

Answer:

$\displaystyle \frac{dy}{d\theta} = \frac{1}{\theta -4} -e^\theta$

Step-by-step explanation:

We are given the function:

$\displaystyle y = \ln \left( \theta - 4 \right) - e^\theta$

And we want to find dy/dθ.

Take the derivative of both sides with respect to θ:

$\displaystyle \frac{dy}{d\theta} = \frac{d}{d\theta}\left[ \ln \left(\theta - 4\right) - e^\theta\right]$

Rewrite:

$\displaystyle \frac{dy}{d\theta} = \frac{d}{d\theta}\left[ \ln\left(\theta -4\right)\right] - \frac{d}{d\theta}\left[e^\theta\right]$

Differentiate. The first differentiation requires the chain rule:

$\displaystyle \begin{aligned}\frac{dy}{d\theta} & = \frac{1}{(\theta - 4)}\cdot \frac{d}{d\theta}\left[ \theta -4\right] - (e^\theta) \\ \\ & = \frac{1}{\theta -4} - e^\theta \end{aligned}$

In conclusion:

$\displaystyle \frac{dy}{d\theta} = \frac{1}{\theta -4} -e^\theta$

Answer 2

Step-by-step explanation:

$y = \ln( \theta - 4) - {e}^{ \theta}$

$\frac{dy}{d \theta} = \frac{d}{d \theta} ( \ln( \theta - 4) ) - \frac{d}{d \theta} ( {e}^{ \theta} )$

$\frac{dy}{d \theta} = \frac{d}{d \theta}( \theta - 4). \frac{1}{\theta - 4} - {e}^{ \theta}$

$\frac{dy}{d \theta} = 1.\frac{1}{ \theta - 4} - {e}^{ \theta}$

$\frac{dy}{d \theta} = \frac{1}{ \theta - 4} - {e}^{ \theta}$