Question

log(1 + e cos theta) differentiate ​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

The differentiation of log(1 + ecosθ) is $\frac{e}{1+ecos\theta}[cos\theta - sin\theta]$.

Step-by-step explanation:

If a single-variable function exists at a given input value, its derivative is defined as the slope of the tangent line to the function's graph at that location. The best linear approximation of the function close to that input value is the tangent line. Due to this, the derivative is frequently referred to as the "instantaneous rate of change," or the ratio of the instantaneous change in the dependent variable to the change in the independent variable.

The given equation is log(1 + ecosθ)

Differentiating w.r.t θ, we get

$\implies y = log(1+ecos\theta)\\\\\implies \frac{dy}{d\theta} = \frac{1}{1+ecos\theta}\frac{d}{d\theta}(1+ecos\theta)\\\\\implies \frac{dy}{d\theta} = \frac{1}{1+ecos\theta}[0 + e(-sin\theta) + cos\theta(e)]\\\\ \implies \frac{dy}{d\theta} = \frac{e}{1+ecos\theta}[cos\theta- sin\theta]$

Hence, the differentiation of log(1 + ecosθ) is $\frac{e}{1+ecos\theta}[cos\theta - sin\theta]$.

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https://brainly.in/question/758525

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