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What is \[\dfrac{d}{{dx}}\left( {\cos x\dfrac{{dy}}{{dz}}} \right)\], where \[z = \sin x\]

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Last updated date: 21st Jul 2024
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Answer
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Hint: Here in this question, we have to find the derivative of the given function, here the function is a trigonometric function. To solve this, we use the standard differentiation formulas of trigonometry functions. The function also contains the product function, by using the product rule we obtain the solution for the given question.

Complete step by step solution:
 In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
Trigonometry is a branch of mathematics that studies relationships between side lengths and angles of triangles. In trigonometry we have six trigonometry ratios namely sine, cosine, tangent, cosecant, secant and cotangent.
Now we consider the given question
\[\dfrac{d}{{dx}}\left( {\cos x\dfrac{{dy}}{{dz}}} \right)\]
We can solve this by two methods
Method 1:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x\dfrac{{dy}}{{dz}}} \right)\]
The function which we have to differentiate is in the form of a product of two functions. So we apply the product rule for it
The product rule is given by \[\dfrac{d}{{dx}}\left( {u.v} \right) = u.\dfrac{{dv}}{{dx}} + v.\dfrac{{du}}{{dx}}\], therefore we have
\[ \Rightarrow \cos x.\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dz}}} \right) + \dfrac{{dy}}{{dz}}.\dfrac{d}{{dx}}(\cos x)\]------- (1)
As we know that \[z = \sin x\], on differentiating z we have
\[ \Rightarrow dz = \cos x.dx\]
Substituting this in the (1)
\[ \Rightarrow \cos x.\dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{\cos x.dx}}} \right) + \dfrac{{dy}}{{\cos x.dx}}.\dfrac{d}{{dx}}(\cos x)\]
Using the reciprocal trigonometric ratios we have
\[ \Rightarrow \cos x.\dfrac{d}{{dx}}\left( {\sec x\dfrac{{dy}}{{dx}}} \right) + \sec x\dfrac{{dy}}{{dx}}.\dfrac{d}{{dx}}(\cos x)\]
On differentiating we have
\[ \Rightarrow \cos x.\left( {\sec x.\dfrac{{{d^2}y}}{{d{x^2}}} + \sec x.\tan x\dfrac{{dy}}{{dx}}} \right) + \sec x\dfrac{{dy}}{{dx}}.( - \sin x)\]
On simplifying the trigonometric ratios.
\[ \Rightarrow \cos x.\sec x.\dfrac{{{d^2}y}}{{d{x^2}}} + \cos x.\sec x.\tan x\dfrac{{dy}}{{dx}} - \sin x\sec x\dfrac{{dy}}{{dx}}\]
\[ \Rightarrow \cos x.\dfrac{1}{{\cos x}}.\dfrac{{{d^2}y}}{{d{x^2}}} + \cos x.\dfrac{1}{{\cos x}}.\dfrac{{\sin x}}{{\cos x}}\dfrac{{dy}}{{dx}} - \sin x.\dfrac{1}{{\cos x}}\dfrac{{dy}}{{dx}}\]
Cancelling the terms which gets cancels, we have
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} + \tan x\dfrac{{dy}}{{dx}} - \tan x\dfrac{{dy}}{{dx}}\]
On further simplifying we have
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}\]
Hence this the derivative.
We can also solve this by another method
Method 2:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x\dfrac{{dy}}{{dz}}} \right)\]-------- (2)
As we know that \[z = \sin x\], on differentiating z we have
\[ \Rightarrow dz = \cos x.dx\]
Substituting this in (2)
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x\dfrac{{dy}}{{\cos x.dx}}} \right)\]
On simplifying we have
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right)\]
On differentiating we have
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}\]
Therefore \[\dfrac{d}{{dx}}\left( {\cos x\dfrac{{dy}}{{dz}}} \right) = \dfrac{{{d^2}y}}{{d{x^2}}}\]
So, the correct answer is “\[\dfrac{{{d^2}y}}{{d{x^2}}}\]”.

Note: We can also solve this by another method
Method 2:
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x\dfrac{{dy}}{{dz}}} \right)\]-------- (2)
As we know that \[z = \sin x\], on differentiating z we have
\[ \Rightarrow dz = \cos x.dx\]
Substituting this in (2)
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x\dfrac{{dy}}{{\cos x.dx}}} \right)\]
On simplifying we have
\[ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{dy}}{{dx}}} \right)\]
On differentiating we have
\[ \Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}}\]
Therefore \[\dfrac{d}{{dx}}\left( {\cos x\dfrac{{dy}}{{dz}}} \right) = \dfrac{{{d^2}y}}{{d{x^2}}}\]