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  • Find the derivative of the function given by   <img alt="f(x)=\frac{\cos \left(b x^2+c x+d\right)}{\sin \left(e x^2+f x+g\right)}" src="https://entrancecorner.oncodecogs.com/gif.latex?f%2

Find the derivative of the function given by   f(x)=\frac{\cos \left(b x^2+c x+d\right)}{\sin \left(e x^2+f x+g\right)}

Option: 1

\begin{aligned} & \frac{\sin \left(e x^2+f x+g\right) *\left(-\sin \left(b x^2+c x+d\right) *(2 b x+c)\right)-\cos \left(b x^2+c x+d\right) *\left(\cos \left(e x^2+f x+g\right) *(2 e x+f)\right)}{\sin \left(e x^2+f x+g\right)^2 9} \\ & \frac{\cos \left(e x^2+f x+g\right) *\left(-\sin \left(b x^2+c x+d\right) *(2 b x+c)\right)-\cos \left(b x^2+c x+d\right) *\left(\cos \left(e x^2+f x+g\right) *(2 e x+f)\right)}{\sin \left(e x^2+f x+g\right)^2 9} \end{aligned}


Option: 2

\frac{\cos \left(e x^2+f x+g\right) *\left(-\sin \left(b x^2+c x+d\right) *(2 b x+c)\right)-\cos \left(b x^2+c x+d\right) *\left(\cos \left(e x^2+f x+g\right) *(2 e x+f)\right)}{\sin \left(e x^2+f x+g\right)^2 9}


Option: 3

\frac{\cos \left(e x^2+f x+g\right) *\left(-\cos \left(b x^2+c x+d\right) *(2 b x+c)\right)-\cos \left(b x^2+c x+d\right) *\left(\cos \left(e x^2+f x+g\right) *(2 e x+f)\right)}{\sin \left(e x^2+f x+g\right)^2 9}


Option: 4

\frac{\sin \left(e x^2+f x+g\right) *\left(-\cos \left(b x^2+c x+d\right) *(2 b x+c)\right)-\sin \left(b x^2+c x+d\right) *\left(\cos \left(e x^2+f x+g\right) *(2 e x+f)\right)}{\cos \left(e x^2+f x+g\right)^2 9}


Answers (1)

best_answer

To find the derivative of the function f(x)=\frac{\cos \left(b x^2+c x+d\right)}{\sin \left(e x^2+f x+g\right)}

use the quotient rule of differentiation. The quotient rule states that if  a function of the form  f(x)=\frac{u(x)}{v(x)}  

where(x)  and v(x) are differentiable functions, the derivative is given by:f^{\prime}(x)=\frac{v(x) * u^{\prime}(x)-u(x) * v^{\prime}(x)}{v(x)^2}

Let's apply the quotient rule to find the derivative of the given function:


\begin{aligned} &u(x)=\cos \left(b x^2+c x+d\right)\\ &v(x)=\sin \left(e x^2+f x+g\right) \end{aligned}

Now, to find the derivatives of u(x) and v(x) 

\begin{aligned} &u^{\prime}(x)=-\sin \left(b x^2+c x+d\right) *(2 b x+c)\\ &v^{\prime}(x)=\cos \left(e x^2+f x+g\right) *(2 e x+f) \end{aligned}

Substituting these derivatives into the quotient rule formula, get:


$$ f^{\prime}(x)=\frac{v(x) * u^{\prime}(x)-u(x) * v^{\prime}(x)}{v(x)^2} $$ $$ \begin{aligned} & =\left[\sin \left(e x^{\wedge} 2+f x+g\right)^*\left(-\sin \left(b x^{\wedge} 2+c x+d\right) *(2 b x+c)\right)-\cos \left(b x^{\wedge} 2+c x+d\right) *\left(\cos \left(e x^{\wedge} 2+f x+g\right)^*\right.\right. \\ & (2 e x+f))] /\left[\sin \left(e x^{\wedge} 2+f x+g\right)\right]^{\wedge} 2 \end{aligned} $$ Simplifying further: =\frac{\sin \left(e x^2+f x+g\right) *\left(-\sin \left(b x^2+c x+d\right) *(2 b x+c)\right)-\cos \left(b x^2+c x+d\right) *\left(\cos \left(c x^2+f x+g\right) *(2 c x+f)\right)}{\sin \left(c x^2+f x+g\right)^2 9}

 

Posted by

Sanket Gandhi

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