What is the derivative of y(θ)=ln(θ+1)-e^θ ?

The derivative of y(θ)=ln(θ+1)-e^θ is 1/(θ+1)-e^θ

What is the first derivative of y(θ)=ln(θ+1)-e^θ ?

The first derivative of y(θ)=ln(θ+1)-e^θ is 1/(θ+1)-e^θ

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derivative of

y (θ) = ln (θ + 1) - e^{θ}

\frac{1}{θ + 1} - e^{θ}

Solution steps

\frac{d}{d θ} (ln (θ + 1) - e^{θ})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d θ} (ln (θ + 1)) - \frac{d}{d θ} (e^{θ})

\frac{d}{d θ} (ln (θ + 1)) = \frac{1}{θ + 1}

\frac{d}{d θ} (e^{θ}) = e^{θ}

= \frac{1}{θ + 1} - e^{θ}

\frac{\partial}{\partial x} (ln (x + y) - 5 y)

\frac{d}{d x} ((x + 3)^{x})

\frac{d}{d x} (x^{5 e^{x}})

ln (\frac{x}{2} - 3), [7, 9]

\int_{0}^{3} \int_{0}^{y} e^{- y^{2}} d x d y

What is the derivative of y(θ)=ln(θ+1)-e^θ ?
The derivative of y(θ)=ln(θ+1)-e^θ is 1/(θ+1)-e^θ
What is the first derivative of y(θ)=ln(θ+1)-e^θ ?
The first derivative of y(θ)=ln(θ+1)-e^θ is 1/(θ+1)-e^θ