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y = cosx + sin2x

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Hint: Here, we will use the concept that the derivative of a function which is a sum of two different functions is given as the sum of derivative of both the functions.

Complete step-by-step answer:

We know that the derivative of a function y = f(x) of a variable x is a measure of the rate at which the value of y of the function changes with respect to the change of the variable x. This process is called the derivative of f(x) with respect to x.

We also know that derivatives mainly represent the slope of a graph at any given point. It means that it is a ratio of change in the value of the function to the change in the independent variable.

Now, the function given to us is:

y = cosx + sin2x

On differentiating both sides with respect to x, we get:

$\dfrac{dy}{dx}=\dfrac{d\left( \cos x \right)}{dx}+\dfrac{d\left( \sin 2x \right)}{dx}..........(1)$

We know that the derivative of cosx is = -sinx. So, we have:

$\dfrac{d\left( \cos x \right)}{dx}=-\sin x.........(2)$

Now, to find the derivative of sin2x, we may use chain rule.

Chain rule says that if we have to find the derivative of a function y which is a combination of two functions x and t, then:

$\dfrac{dy}{dt}=\dfrac{dy}{dx}\times \dfrac{dx}{dt}$

So, using this chain rule for sin2x and putting 2x =t, we have:

$\begin{align}

& \dfrac{d\left( \sin 2x \right)}{dx}=\dfrac{d\left( \sin t \right)}{dx} \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{d\left( \sin t \right)}{dt}\times \dfrac{dt}{dx} \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\cos t\times \dfrac{d\left( 2x \right)}{dx} \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\cos 2x....................(2) \\

\end{align}$

On substituting the values from equation (2) and (3) in equation (1), we get:

$\dfrac{dy}{dx}=-\sin x+2\cos 2x$

Hence, the derivative of function y = cosx + sin2x is equal to –sinx + 2cos2x.

Note: Students should note here that we apply chain rule to find the derivative of sin2x because it is a function which is a combination of two functions. Do not get confused with sine and cosine derivatives.

Complete step-by-step answer:

We know that the derivative of a function y = f(x) of a variable x is a measure of the rate at which the value of y of the function changes with respect to the change of the variable x. This process is called the derivative of f(x) with respect to x.

We also know that derivatives mainly represent the slope of a graph at any given point. It means that it is a ratio of change in the value of the function to the change in the independent variable.

Now, the function given to us is:

y = cosx + sin2x

On differentiating both sides with respect to x, we get:

$\dfrac{dy}{dx}=\dfrac{d\left( \cos x \right)}{dx}+\dfrac{d\left( \sin 2x \right)}{dx}..........(1)$

We know that the derivative of cosx is = -sinx. So, we have:

$\dfrac{d\left( \cos x \right)}{dx}=-\sin x.........(2)$

Now, to find the derivative of sin2x, we may use chain rule.

Chain rule says that if we have to find the derivative of a function y which is a combination of two functions x and t, then:

$\dfrac{dy}{dt}=\dfrac{dy}{dx}\times \dfrac{dx}{dt}$

So, using this chain rule for sin2x and putting 2x =t, we have:

$\begin{align}

& \dfrac{d\left( \sin 2x \right)}{dx}=\dfrac{d\left( \sin t \right)}{dx} \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\dfrac{d\left( \sin t \right)}{dt}\times \dfrac{dt}{dx} \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\cos t\times \dfrac{d\left( 2x \right)}{dx} \\

& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=2\cos 2x....................(2) \\

\end{align}$

On substituting the values from equation (2) and (3) in equation (1), we get:

$\dfrac{dy}{dx}=-\sin x+2\cos 2x$

Hence, the derivative of function y = cosx + sin2x is equal to –sinx + 2cos2x.

Note: Students should note here that we apply chain rule to find the derivative of sin2x because it is a function which is a combination of two functions. Do not get confused with sine and cosine derivatives.

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