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Question: How do you find the derivative of \((\cos x)\)using the limit definition ?...

How do you find the derivative of (cos⁡x)(\cos x)using the limit definition ?

Explanation

Solution

Hint : In order to find the first derivative of the above expression using the limit definition , we need to solve it step by step . Also we need to know some important concepts of the terms used in the question which will help us in solving the question , So, the two terms Derivatives and Limits are used and we should have knowledge about the same before solving the question . So , Limit is defined as a value that a function approaches as the input, and it produces some value . We will be using some limited formula and properties required to solve the question . And the derivative refers to the instantaneous rate of change of a quantity with respect to the other , The derivative of a function is represented in the below-given formula. lim⁡h→0f(x+h)−f(x)h\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}.

Complete step-by-step answer :
Given a function (cos⁡x)(\cos x) , we know the formula of the derivative lim⁡h→0f(x+h)−f(x)h\mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}
So, f(x)=cos⁡xf(x) = \cos x
Applying limits and derivative to the function , we get –
f′(x)=lim⁡h→0cos⁡(x+h)−cos⁡(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x + h) - \cos (x)}}{h}
Now we will use the trigonometry formula cos⁡(x+h)=cos⁡(x)cos⁡(h)−sin⁡(x)sin⁡(h)\cos (x + h) = \cos (x)\cos (h) - \sin (x)\sin (h) to rewrite the derivative of (cos⁡x)(\cos x) as –
f′(x)=lim⁡h→0cos⁡(x)cos⁡(h)−sin⁡(x)sin⁡(h)−cos⁡(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x)\cos (h) - \sin (x)\sin (h) - \cos (x)}}{h}

Rewrite as follows –
f′(x)=lim⁡h→0cos⁡(x)cos⁡(h)−sin⁡(x)sin⁡(h)−cos⁡(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x)\cos (h) - \sin (x)\sin (h) - \cos (x)}}{h}
f′(x)=lim⁡h→0cos⁡(x)(cos⁡(h)−1)−sin⁡(x)sin⁡(h)−cos⁡(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x)(\cos (h) - 1) - \sin (x)\sin (h) - \cos (x)}}{h}
Use the theorem on limits that states: the limit of a difference of two functions is equal to the difference of the limits, to rewrite as follows –
f′(x)f′(x)=lim⁡h→0cos⁡(x)(cos⁡(h)−1)h−lim⁡h→0sin⁡(x)sin⁡(h)hf'(x)f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x)(\cos (h) - 1)}}{h} - \mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (x)\sin (h)}}{h}----------------equation 1
Now there are some fundamental trigonometric limits :
lim⁡θ→0sin⁡θθ=1\mathop {\lim }\limits_{\theta \to 0} \dfrac{{\sin \theta }}{\theta } = 1 and lim⁡θ→0cos⁡θ−1θ=0\mathop {\lim }\limits_{\theta \to 0} \dfrac{{\cos \theta - 1}}{\theta } = 0
Applying these limits in equation 1 we get –

f′(x)=cos⁡(x)(lim⁡h→0(cos⁡(h)−1)h)−sin⁡(x)(lim⁡h→0sin⁡(h)h) f′(x)=cos⁡(x)(0)−sin⁡(x)(1) f′(x)=0−sin⁡(x) f′(x)=−sin⁡(x)  f'(x) = \cos (x)\left( {\mathop {\lim }\limits_{h \to 0} \dfrac{{(\cos (h) - 1)}}{h}} \right) - \sin (x)\left( {\mathop {\lim }\limits_{h \to 0} \dfrac{{\sin (h)}}{h}} \right) \\\ f'(x) = \cos (x)(0) - \sin (x)(1) \\\ f'(x) = 0 - \sin (x) \\\ f'(x) = - \sin (x) \\\

Therefore , the derivative of (cos⁡x)(\cos x)using the limit definition is f′(x)=−sin⁡(x)f'(x) = - \sin (x)
f(x)=cos⁡xf(x) = \cos x
cos⁡′(x)=−sin⁡(x)\cos '(x) = - \sin (x)
So, the correct answer is “−sin⁡(x)- \sin (x)”.

Note : Always try to understand the mathematical statement carefully and keep things distinct .
Remember the properties and apply appropriately .
Choose the options wisely , it's better to break the question and then solve part by part .
Cross check the answer and always keep the final answer simplified .
Avoid jumping the steps as it can create an error. To solve the problem, ideas about limits are utmost important.