Question

How do you differentiate $y=4\ln x+2\cos x-3{{e}^{x}}$?

Answer 1

Hint: Now to differentiate the given function we will use the property of differentiation which states $\dfrac{d\left( f\pm g \right)}{dx}=\dfrac{d\left( f \right)}{dx}\pm \dfrac{d\left( g \right)}{dx}$ . Now we will use the property $\dfrac{d\left( cf \right)}{dx}=c\dfrac{df}{dx}$ and take constant out of all the terms. Now we know that the differentiation of $\cos x$ is given by $-\sin x$ , the differentiation of ${{e}^{x}}$ is given by ${{e}^{x}}$and the differentiation of $\ln x=\dfrac{1}{x}$ . Hence using this we get the differentiation of the given function.

Complete step-by-step answer:
Now consider the given function $y=4\ln x+2\cos x-3{{e}^{x}}$ .
To differentiate the function we will find the value of $\dfrac{dy}{dx}$ .
Now to differentiate the given function let us first understand some properties of differentiation.
Now we know that the differentiation of addition or subtraction of two function is given by $\Rightarrow \dfrac{d\left( f+g \right)}{dx}=\dfrac{d\left( f \right)}{dx}+\dfrac{d\left( g \right)}{dx}$ and $\dfrac{d\left( f-g \right)}{dx}=\dfrac{d\left( f \right)}{dx}-\dfrac{d\left( g \right)}{dx}$
Hence using this we can write the given differential as
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d\left( 4\ln x \right)}{dx}+\dfrac{d\left( 2\cos x \right)}{dx}-\dfrac{d\left( 3{{e}^{x}} \right)}{dx}\]
Now we know that for differentiation of any function f $\dfrac{d\left( cf \right)}{dx}=c\dfrac{df}{dx}$ where c is any constant.
Hence in the above equation we will take out the respective constants in each term.
$\Rightarrow \dfrac{dy}{dx}=4\dfrac{d\left( \ln x \right)}{dx}+2\dfrac{d\left( \cos x \right)}{dx}-3\dfrac{d\left( {{e}^{x}} \right)}{dx}$
Now we have the differentiation of $\cos x$ is given by $-\sin x$, the differentiation of $\ln x$ is $\dfrac{1}{x}$ and the differentiation of ${{e}^{x}}$ is given by ${{e}^{x}}$ .
Hence using this values of differentiation we get,
$\Rightarrow \dfrac{dy}{dx}=\dfrac{4}{x}-2\sin x-3{{e}^{x}}$
Hence the differentiation of the given function is $\dfrac{4}{x}-2\sin x-3{{e}^{x}}$

Note: Now note that the differentiation we have the differentiation of the sum of functions is sum of differentiation of function. But note that this is not the same case with multiplication and division. For multiplication we have $\dfrac{d\left( f.g \right)}{dx}=g\dfrac{df}{dx}+f\dfrac{dg}{dx}$ . Similarly of we have for function in division we have the differentiation of the function as $\left( \dfrac{f}{g} \right)'=\dfrac{f'g-g'f}{{{g}^{2}}}$ . Also we can use multiplication rule to find the differentiation of fractions by considering $\dfrac{df}{dx}.\dfrac{d\left( \dfrac{1}{g} \right)}{dx}$ .