Question

Find the derivative with respect to \(x\) of the function y specified implicitly by \(\int_{0}^{y} e^{t} d t+\int_{0}^{x} \cos t d t=0\)

Short answer

Question: Find the derivative of the function y with respect to x, given the equation \(\int_{0}^{y} e^{t} dt + \int_{0}^{x} \cos t dt = 0\). Answer: \(\frac{dy}{dx} = \frac{-\cos{x}}{e^{y}}\)

Step-by-step solution

Differentiate both sides with respect to x

Using the chain rule for differentiation, we differentiate the given equation: \(\frac{d}{dx} \left(\int_{0}^{y} e^{t} dt\right) + \frac{d}{dx} \left(\int_{0}^{x} \cos t dt\right) = \frac{d}{dx} (0)\)

Apply the Fundamental Theorem of Calculus to differentiate the integral expressions

We have: \(\frac{d}{dx} \left(\int_{0}^{y} e^{t} dt\right) = e^{y} \frac{dy}{dx}\) and \(\frac{d}{dx} \left(\int_{0}^{x} \cos t dt\right) = \cos{x}\) So, the differentiated equation becomes: \(e^{y}\frac{dy}{dx}+\cos{x} = 0\)

Solve for \(\frac{dy}{dx}\)

Lastly, we isolate \(\frac{dy}{dx}\): \(\frac{dy}{dx}=\frac{-\cos{x}}{e^{y}}\) The derivative of the function y with respect to x is: \(\frac{dy}{dx}=\frac{-\cos{x}}{e^{y}}\)