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$\cos (\sin x)$

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Verified

Hint: In $\cos (\sin x)$Differentiate$\cos $ and then the bracket i.e. $\sin x$ .

So in general we want to find $\dfrac{d\cos (\sin x)}{dx}$

So we know $\dfrac{d\sin x}{dx}=\cos x$ and that of $\dfrac{d\cos x}{dx}=-\sin x$

So now explicitly chain rule of differentiation,

chain rule of differentiation is $f(g(x))={{f}^{'}}(g(x)){{g}^{'}}(x)$,

The $y$ in the formula for the derivative is the price we pay for not making the function explicit. It replaces the explicit form of the function, whatever that may be.

An explicit function is a function in which one variable is defined only in terms of the other variable.

An explicit function is one which is given in terms of the independent variable.

$y=-\dfrac{3}{5}x+\dfrac{7}{5}$ gives $y$ explicitly as a function of $x$.

So the Examples of explicit functions are $y={{x}^{2}}+7$,

So $y$ is the dependent variable and is given in terms of the independent variable $x$.

Note that $y$ is the subject of the formula.

Whereas implicit functions are usually given in terms of both dependent and independent

variables.

So the Examples of implicit functions are $y+{{x}^{2}}+2x=0$,

Sometimes it is not convenient to express function explicitly,

For example circle ${{x}^{2}}+{{y}^{2}}=4$

It is often easier to differentiate an implicit function without having to rearrange it, by differentiating each term in turn.

So it could be written as,

$y=\sqrt{4-{{x}^{2}}}$and$y=-\sqrt{4-{{x}^{2}}}$

So now differentiating, We get,

$\begin{align}

& =\dfrac{d\cos (\sin x)}{dx} \\

& =-\sin (\sin x)\dfrac{d\sin x}{dx} \\

\end{align}$

In above we can see that we had differentiated the function, in such cases we can take

example,

$\dfrac{d}{dx}\cos ({{x}^{2}})=-\sin ({{x}^{2}})\dfrac{d({{x}^{2}})}{dx}=-2x\sin ({{x}^{2}})$

So we had differentiated first outer part and then inner bracket,

So applying the same in above problem, we get,

So Above problem becomes,

$\begin{align}

& =-\sin (\sin x)\dfrac{d\sin x}{dx} \\

& =-\cos x\sin (\sin x) \\

\end{align}$

So we get the differentiation of $\cos (\sin x)$ that is we get the final answer as,

$\dfrac{d\cos (\sin x)}{dx}=-\cos x\sin (\sin x)$

So in this way we can differentiate.

Note: So now you must be aware of what you have to differentiate.

Generally what happens the student makes mistakes in differentiation of $\sin x$and $\cos

x$. So be careful with that . Remember that $\dfrac{d\sin x}{dx}=\cos x$and $\dfrac{d\cos

x}{dx}=-\sin x$.

While solving above problem mostly students forget to differentiate the bracket such as instead of

this$\dfrac{d}{dx}\cos ({{x}^{2}})=-\sin ({{x}^{2}})\dfrac{d({{x}^{2}})}{dx}=-2x\sin ({{x}^{2}})$Student write it as $\dfrac{d}{dx}\cos ({{x}^{2}})=-\sin ({{x}^{2}})$. So this should

be avoided.

So in general we want to find $\dfrac{d\cos (\sin x)}{dx}$

So we know $\dfrac{d\sin x}{dx}=\cos x$ and that of $\dfrac{d\cos x}{dx}=-\sin x$

So now explicitly chain rule of differentiation,

chain rule of differentiation is $f(g(x))={{f}^{'}}(g(x)){{g}^{'}}(x)$,

The $y$ in the formula for the derivative is the price we pay for not making the function explicit. It replaces the explicit form of the function, whatever that may be.

An explicit function is a function in which one variable is defined only in terms of the other variable.

An explicit function is one which is given in terms of the independent variable.

$y=-\dfrac{3}{5}x+\dfrac{7}{5}$ gives $y$ explicitly as a function of $x$.

So the Examples of explicit functions are $y={{x}^{2}}+7$,

So $y$ is the dependent variable and is given in terms of the independent variable $x$.

Note that $y$ is the subject of the formula.

Whereas implicit functions are usually given in terms of both dependent and independent

variables.

So the Examples of implicit functions are $y+{{x}^{2}}+2x=0$,

Sometimes it is not convenient to express function explicitly,

For example circle ${{x}^{2}}+{{y}^{2}}=4$

It is often easier to differentiate an implicit function without having to rearrange it, by differentiating each term in turn.

So it could be written as,

$y=\sqrt{4-{{x}^{2}}}$and$y=-\sqrt{4-{{x}^{2}}}$

So now differentiating, We get,

$\begin{align}

& =\dfrac{d\cos (\sin x)}{dx} \\

& =-\sin (\sin x)\dfrac{d\sin x}{dx} \\

\end{align}$

In above we can see that we had differentiated the function, in such cases we can take

example,

$\dfrac{d}{dx}\cos ({{x}^{2}})=-\sin ({{x}^{2}})\dfrac{d({{x}^{2}})}{dx}=-2x\sin ({{x}^{2}})$

So we had differentiated first outer part and then inner bracket,

So applying the same in above problem, we get,

So Above problem becomes,

$\begin{align}

& =-\sin (\sin x)\dfrac{d\sin x}{dx} \\

& =-\cos x\sin (\sin x) \\

\end{align}$

So we get the differentiation of $\cos (\sin x)$ that is we get the final answer as,

$\dfrac{d\cos (\sin x)}{dx}=-\cos x\sin (\sin x)$

So in this way we can differentiate.

Note: So now you must be aware of what you have to differentiate.

Generally what happens the student makes mistakes in differentiation of $\sin x$and $\cos

x$. So be careful with that . Remember that $\dfrac{d\sin x}{dx}=\cos x$and $\dfrac{d\cos

x}{dx}=-\sin x$.

While solving above problem mostly students forget to differentiate the bracket such as instead of

this$\dfrac{d}{dx}\cos ({{x}^{2}})=-\sin ({{x}^{2}})\dfrac{d({{x}^{2}})}{dx}=-2x\sin ({{x}^{2}})$Student write it as $\dfrac{d}{dx}\cos ({{x}^{2}})=-\sin ({{x}^{2}})$. So this should

be avoided.

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