How do you differentiate $ y = \cos (x) $ from the first principles?
Answer
583.5k+ views
Hint: Start by considering $ f(x) $ as the function of $ x $ . Next step is substitution. Substitute the values in place of the terms to make the equation easier to solve. Then by definition of the derivative we use the following formula: $ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} $
Complete step-by-step answer:
First we will start off by directly applying the definition of the derivative.
$ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} $
So, now with $ f(x) = \cos x $ here we have,
$ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x + h) - \sin x}}{h} $
Now we will be using $ \cos (A + B) = \cos A\cos B - \sin A\sin B $ so we will get
$
f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x\cosh h - \sinh \sin x - \cos x}}{h} \\
f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x(\cosh h - 1) - \sinh \sin x}}{h} \\
f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x(\cos h - 1)}}{h} - \mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh \sin x}}{h} \\
f'(x) = (\cos x)\mathop {\lim }\limits_{h \to 0} \dfrac{{(\cos h - 1)}}{h} - (\sin x)\mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h} \;
$
Now here we have to rely on some standard limits such as:
$ \mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h} = 1 $ and $ \mathop {\lim }\limits_{h \to 0} \dfrac{{\cosh - 1}}{h} = 0 $ .
So, now we have:
$
f'(x) = 0 - (\sin x)(1) \\
= - \sin x \;
$
Hence, the derivative of $ f(x) = \cos (x) $ from the first principles will be $ - \sin x $ .
So, the correct answer is “-sin x”.
Note: A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behaviour of the original system under diverse conditions.
Complete step-by-step answer:
First we will start off by directly applying the definition of the derivative.
$ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} $
So, now with $ f(x) = \cos x $ here we have,
$ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos (x + h) - \sin x}}{h} $
Now we will be using $ \cos (A + B) = \cos A\cos B - \sin A\sin B $ so we will get
$
f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x\cosh h - \sinh \sin x - \cos x}}{h} \\
f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x(\cosh h - 1) - \sinh \sin x}}{h} \\
f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\cos x(\cos h - 1)}}{h} - \mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh \sin x}}{h} \\
f'(x) = (\cos x)\mathop {\lim }\limits_{h \to 0} \dfrac{{(\cos h - 1)}}{h} - (\sin x)\mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h} \;
$
Now here we have to rely on some standard limits such as:
$ \mathop {\lim }\limits_{h \to 0} \dfrac{{\sinh }}{h} = 1 $ and $ \mathop {\lim }\limits_{h \to 0} \dfrac{{\cosh - 1}}{h} = 0 $ .
So, now we have:
$
f'(x) = 0 - (\sin x)(1) \\
= - \sin x \;
$
Hence, the derivative of $ f(x) = \cos (x) $ from the first principles will be $ - \sin x $ .
So, the correct answer is “-sin x”.
Note: A derivative is the rate of change of a function with respect to a variable. Derivatives are fundamental to the solution of problems in calculus and differential equations. In general, scientists observe changing systems to obtain rate of change of some variable of interest, incorporate this information into some differential equation, and use integration techniques to obtain a function that can be used to predict the behaviour of the original system under diverse conditions.
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