Question:
Differentiate the following w.r.t. \(x\):
\(cosx.cos2x.cos3x\)
Differentiate the following w.r.t. \(x\):
\(cosx.cos2x.cos3x\)
\(cosx.cos2x.cos3x\)
Updated On: Jun 7, 2025
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Solution and Explanation
The correct answer is \(∴\frac{dy}{dx}=-cos\,x.cos\,2x.cos\,3x[tan\,x+2tan\,2x+3tan\,3x]\)
Let \(y=cosx.cos2x.cos3x\)
Taking logarithm on both the sides,we obtain
\(log\,y=log(cos\,x.cos\,2x.cos\,3x)\)
\(⇒log\,y=log(cos\,x)+log(cos\,2x)+log(cos\,3x)\)
Differentiating both sides with respect to x,we obtain
\(\frac{1}{y}\frac{dy}{dx}=\frac{1}{cos\,x}.\frac{d}{dx}(cos\,x)+\frac{1}{cos\,2x}.\frac{d}{dx}(cos\,2x)+\frac{1}{cos\,3x}.\frac{d}{dx}(cos\,3x)\)
\(⇒\frac{dy}{dx}=y[-\frac{sin\,x}{cos\,x}-\frac{sin\,2x}{cos\,2x}.\frac{d}{dx}(2x)-\frac{sin\,3x}{cos\,3x}.\frac{d}{dx}(3x)]\)
\(∴\frac{dy}{dx}=-cos\,x.cos\,2x.cos\,3x[tan\,x+2tan\,2x+3tan\,3x]\)
Let \(y=cosx.cos2x.cos3x\)
Taking logarithm on both the sides,we obtain
\(log\,y=log(cos\,x.cos\,2x.cos\,3x)\)
\(⇒log\,y=log(cos\,x)+log(cos\,2x)+log(cos\,3x)\)
Differentiating both sides with respect to x,we obtain
\(\frac{1}{y}\frac{dy}{dx}=\frac{1}{cos\,x}.\frac{d}{dx}(cos\,x)+\frac{1}{cos\,2x}.\frac{d}{dx}(cos\,2x)+\frac{1}{cos\,3x}.\frac{d}{dx}(cos\,3x)\)
\(⇒\frac{dy}{dx}=y[-\frac{sin\,x}{cos\,x}-\frac{sin\,2x}{cos\,2x}.\frac{d}{dx}(2x)-\frac{sin\,3x}{cos\,3x}.\frac{d}{dx}(3x)]\)
\(∴\frac{dy}{dx}=-cos\,x.cos\,2x.cos\,3x[tan\,x+2tan\,2x+3tan\,3x]\)
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Logarithmic Differentiation
Logarithmic differentiation is a method to find the derivatives of some complicated functions, using logarithms. There are cases in which differentiating the logarithm of a given function is simpler as compared to differentiating the function itself. By the proper usage of properties of logarithms and chain rule finding, the derivatives become easy. This concept is applicable to nearly all the non-zero functions which are differentiable in nature.
Therefore, in calculus, the differentiation of some complex functions is done by taking logarithms and then the logarithmic derivative is utilized to solve such a function.