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QExample - 7: HC Verma: Concepts of Physics - Volume 1 Physics and Mathematics Example - 7 Solution: Find {dy}{dx} if y=e^{x}sin~x

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often mistakenly apply the chain rule or attempt to differentiate the product as a single composite function, or they fail to maintain the correct order in the product rule formula $u'v + uv'$, leading to sign errors or missing terms. Always explicitly identify $u$ and $v$ and apply the rule strictly as the sum of the derivative of the first times the second plus the first times the derivative of the second.

Find

dydx\frac{dy}{dx}
if
y=exsin xy=e^{x}sin~x

Step-by-Step Verified Solution

\textbf{Problem:}


Find dydx\frac{dy}{dx} if y=exsin⁡xy = e^x \sin x.


\textbf{Solution:}


We need to use the product rule for differentiation. The product rule states that if y=u(x)v(x)y = u(x)v(x), then:


dydx=u′(x)v(x)+u(x)v′(x)\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)


In our case, we have:


* u(x)=exu(x) = e^x

* v(x)=sin⁡xv(x) = \sin x


Now, we find the derivatives of u(x)u(x) and v(x)v(x):


* u′(x)=ddx(ex)=exu'(x) = \frac{d}{dx}(e^x) = e^x

* v′(x)=ddx(sin⁡x)=cos⁡xv'(x) = \frac{d}{dx}(\sin x) = \cos x


Now, we apply the product rule:


dydx=u′(x)v(x)+u(x)v′(x)\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)


dydx=(ex)(sin⁡x)+(ex)(cos⁡x)\frac{dy}{dx} = (e^x)(\sin x) + (e^x)(\cos x)


We can factor out exe^x:


dydx=ex(sin⁡x+cos⁡x)\frac{dy}{dx} = e^x (\sin x + \cos x)


\textbf{Therefore, the derivative is:}


dydx=ex(sin⁡x+cos⁡x)\frac{dy}{dx} = e^x (\sin x + \cos x)

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