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Question

Question: How do you find the derivative of \(y=\cos \left( 4{{x}^{3}} \right)\)?...

How do you find the derivative of y=cos(4x3)y=\cos \left( 4{{x}^{3}} \right)?

Explanation

Solution

To get the derivative of y=cos(4x3)y=\cos \left( 4{{x}^{3}} \right) with respect to xx . Firstly, suppose θ=4x3\theta =4{{x}^{3}} and get the derivative of θ\theta with respect to xx . Now we can write y=cos(4x3)y=\cos \left( 4{{x}^{3}} \right) as y=cos(θ)y=\cos \left( \theta \right) and after that get the derivative with respect to θ\theta . After combining both the derivative as dydx=dydθ×dθdx\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{d\theta }{dx} we can get the derivative of y=cos(4x3)y=\cos \left( 4{{x}^{3}} \right) with respect to xx .

Complete step by step solution:
y=cos(4x3)y=\cos \left( 4{{x}^{3}} \right) is the given equation in the question.
Since, we are not able to derive the given equation directly. So, we can consider 4x34{{x}^{3}}as:
θ=4x3\theta =4{{x}^{3}}(i)\left( i \right)
Now, we can derive the above equation (i)\left( i \right) with respect to xx as:
dθdx=d(4x3)dx\Rightarrow \dfrac{d\theta }{dx}=\dfrac{d\left( 4{{x}^{3}} \right)}{dx}
Since, numbers are constant in any derivative, so we cannot derive 44. For x3{{x}^{3}} , the derivative is 3x23{{x}^{2}} .
So, the derivation can be written as:
dθdx=4×3x2\Rightarrow \dfrac{d\theta }{dx}=4\times 3{{x}^{2}}
After simplifying the derivation, it would be as:
dθdx=12x2\Rightarrow \dfrac{d\theta }{dx}=12{{x}^{2}}(ii)\left( ii \right)
After using equation (i)\left( i \right), we can write the given equation y=cos(4x3)y=\cos \left( 4{{x}^{3}} \right) as-
y=cos(θ)\Rightarrow y=\cos \left( \theta \right)
Since, The derivative of cos(θ)\cos \left( \theta \right) with respect to θ\theta is sin(θ)-\sin \left( \theta \right) . After derivative the above equation with respect to θ\theta , we will have:
dydθ=sin(θ)\Rightarrow \dfrac{dy}{d\theta }=-\sin \left( \theta \right)
Now, with the use of equation (i)\left( i \right), we can write the above derivative in term of xx as-
dydθ=sin(4x3)\Rightarrow \dfrac{dy}{d\theta }=-\sin \left( 4{{x}^{3}} \right)(iii)\left( iii \right)
For getting the derivative of the given equation y=cos(4x3)y=\cos \left( 4{{x}^{3}} \right) with respect to xx, we will use the following formula:
dydx=dydθ×dθdx\dfrac{dy}{dx}=\dfrac{dy}{d\theta }\times \dfrac{d\theta }{dx}
Using equation (ii)\left( ii \right) and (iii)\left( iii \right) in above formula, we will get:
dydx=sin(4x3)×12x2\Rightarrow \dfrac{dy}{dx}=-\sin \left( 4{{x}^{3}} \right)\times 12{{x}^{2}}
After simplification, we can write the above equation as:
dydx=12x2sin(4x3)\Rightarrow \dfrac{dy}{dx}=-12{{x}^{2}}\sin \left( 4{{x}^{3}} \right)
Hence, the derivative of the given equation y=cos(4x3)y=\cos \left( 4{{x}^{3}} \right) is 12x2sin(4x3)-12{{x}^{2}}\sin \left( 4{{x}^{3}} \right) .

Note:
Here we can check whether the derivative of the given equation is correct or not in the following way-
From the solution, we have:
dydx=12x2sin(4x3)\dfrac{dy}{dx}=-12{{x}^{2}}\sin \left( 4{{x}^{3}} \right)
We can write it as-
dy=12x2sin(4x3)dx\Rightarrow dy=-12{{x}^{2}}\sin \left( 4{{x}^{3}} \right)dx
After applying the symbol of integration both sides:
dy=[12x2sin(4x3)]dx\Rightarrow \int{{}}dy=\int{\left[ -12{{x}^{2}}\sin \left( 4{{x}^{3}} \right) \right]}dx
After integrating the above equation, we will get:
y=12x23x2[cos(4x3)] y=4[cos(4x3)] y=4cos(4x3) \begin{aligned} & \Rightarrow y=-\dfrac{12{{x}^{2}}}{3{{x}^{2}}}\left[ -\cos \left( 4{{x}^{3}} \right) \right] \\\ & \Rightarrow y=-4\left[ -\cos \left( 4{{x}^{3}} \right) \right] \\\ & \Rightarrow y=4\cos \left( 4{{x}^{3}} \right) \\\ \end{aligned}
Now, we got the given equation of the question from the integration of the solution. Hence, the solution is correct.