
Differentiate $ f(x) = x\cos x $ using the product rule
Answer
562.8k+ views
Hint: Only if you know the calculus, you are able to solve this problem. Here, we have to use the product rule, which is also called as ‘ $ uv $ ’ method, where we keep $ u $ as constant and differentiate $ v $ , then keep $ v $ as constant and differentiate $ u $ and finally add both. That’s it...!
Complete step-by-step answer:
Knowing calculus is the primary important in solving many complicated problems in Mathematics. To solve the complex problems, we have to be thorough with the basics, so solve more basic problems like this to master calculus.
Let’s consider the given equation,
$ f(x) = x\cos x $
Let’s consider the ‘ $ uv $ ’ method, take $ u = x $ and $ v = \cos x $ , the formula is $ uv = uv' + vu' $ , where $ u' $ and $ v' $ are the derivatives of $ u $ and $ v $ . Now let’s consider $ uv' $ , which implies,
$ x\dfrac{{d}}{{dx}}(cos x) = x( - \sin x) $ $ (1) $ … (1)
In equation (1) we Kept $ x $ as constant and differentiated $ \cos x $ and when we took the derivative of $ \cos x $ , we got $ ( - \sin x) $ and then we differentiated $ x $ and we got $ (1) $ . Now considering $ vu' $ , which implies,
$ cos x \dfrac{{d}}{{dx}}(x) = \cos x(1) $ … (2)
In equation (2), we kept $ \cos x $ as constant and differentiated $ x $ , and hence we got was $ \cos x $ $ (1) $ ,
Now adding (1) and (2), we get,
$ \dfrac{{d}}{{dx}}(x\cos x )= - x\sin x + \cos x $
Rearranging the terms, we get,
$ \dfrac{{d}}{{dx}}(x\cos x ) = \cos x - x\sin x $
Hence this is our required solution.
So, the correct answer is “$ \cos x - x\sin x $”.
Note: Whenever there is a “ $ ' $ ”in the superscript of a letter it means, it is in the differential form. Mostly we use $ \dfrac{{d}}{{dx}} $ to represent a differential equation, but when we have some complex equation we represent as $ u' $ , in case $ u $ is in the differential form. Before learning calculus, be thorough with the concept called functions, calculus is the application of the function. So, to build your basic level, you must know What a function means.
Complete step-by-step answer:
Knowing calculus is the primary important in solving many complicated problems in Mathematics. To solve the complex problems, we have to be thorough with the basics, so solve more basic problems like this to master calculus.
Let’s consider the given equation,
$ f(x) = x\cos x $
Let’s consider the ‘ $ uv $ ’ method, take $ u = x $ and $ v = \cos x $ , the formula is $ uv = uv' + vu' $ , where $ u' $ and $ v' $ are the derivatives of $ u $ and $ v $ . Now let’s consider $ uv' $ , which implies,
$ x\dfrac{{d}}{{dx}}(cos x) = x( - \sin x) $ $ (1) $ … (1)
In equation (1) we Kept $ x $ as constant and differentiated $ \cos x $ and when we took the derivative of $ \cos x $ , we got $ ( - \sin x) $ and then we differentiated $ x $ and we got $ (1) $ . Now considering $ vu' $ , which implies,
$ cos x \dfrac{{d}}{{dx}}(x) = \cos x(1) $ … (2)
In equation (2), we kept $ \cos x $ as constant and differentiated $ x $ , and hence we got was $ \cos x $ $ (1) $ ,
Now adding (1) and (2), we get,
$ \dfrac{{d}}{{dx}}(x\cos x )= - x\sin x + \cos x $
Rearranging the terms, we get,
$ \dfrac{{d}}{{dx}}(x\cos x ) = \cos x - x\sin x $
Hence this is our required solution.
So, the correct answer is “$ \cos x - x\sin x $”.
Note: Whenever there is a “ $ ' $ ”in the superscript of a letter it means, it is in the differential form. Mostly we use $ \dfrac{{d}}{{dx}} $ to represent a differential equation, but when we have some complex equation we represent as $ u' $ , in case $ u $ is in the differential form. Before learning calculus, be thorough with the concept called functions, calculus is the application of the function. So, to build your basic level, you must know What a function means.
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