Question

find dy/dx , y=cos(sin(2x³-2x+4)

pls answer crctly ​

Answer 1

Solution :

Function of $\sf{y = cos[sin(2x^{3} - 2x + 4)]}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[cos(sin(2x^{3} - 2x + 4))]}{dx}}$

$\textsf{Now by applying the chain rule of differentiation, we get :} \\ \\ \underline{\sf{\bigstar\:Chain\: rule\:of \: differentiation :-}} \\ \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}} \\ \\ \textsf{Taking here,} \\ \bullet\:\sf{The\:value\:of\:y = cos(sin(2x^{3} - 2x + 4))} \\ \bullet\:\sf{The\:value\:of\:u = sin(2x^{3} - 2x + 4)}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[cos(sin(2x^{3} - 2x + 4))]}{d[sin(2x^{3} - 2x + 4)]}\cdot\dfrac{d[sin(2x^{3} - 2x + 4)]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = -sin[sin(2x^{3} - 2x + 4)]\cdot\dfrac{d[sin(2x^{3} - 2x + 4)]}{dx}}$

$\sf{Now,\:by\:differentiating,\:sin(2x^{3} - 2x + 4), w.r.t x,\:we\:get :}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[sin(2x^{3} - 2x + 4)]}{dx}}$

$\textsf{Now by applying the chain rule of differentiation, we get :} \\ \\ \underline{\sf{\bigstar\:Chain\: rule\:of \: differentiation :-}} \\ \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}} \\ \\ \textsf{Taking here,} \\ \bullet\:\sf{The\:value\:of\:y = sin(2x^{3} - 2x + 4)} \\ \bullet\:\sf{The\:value\:of\:u = 2x^{3} - 2x + 4}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[sin(2x^{3} - 2x + 4)]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[sin(2x^{3} - 2x + 4)]}{d(2x^{3} - 2x + 4)}\cdot\dfrac{d(2x^{3} - 2x + 4)}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = cos(2x^{3} - 2x + 4)\cdot\dfrac{d(2x^{3} - 2x + 4)}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = cos(2x^{3} - 2x + 4)\cdot\bigg[\dfrac{d(2x^{3})}{dx} - \dfrac{d(2x)}{dx} + \dfrac{d(4)}{dx}\bigg]}$

$:\implies \sf{\dfrac{dy}{dx} = cos(2x^{3} - 2x + 4)\cdot[6x^{2} - 2 + 0]}$

$:\implies \sf{\dfrac{dy}{dx} = cos(2x^{3} - 2x + 4)\cdot(6x^{2} - 2)}$

$\therefore \sf{\dfrac{dy}{dx} = cos(2x^{3} - 2x + 4)\cdot(6x^{2} - 2)}$

$\textsf{Now, by substituting the derivative of}\: \\ \sf sin(2x^{3} - 2x + 4)\:in\: the\: Equation, we \:get :$

$:\implies \sf{\dfrac{dy}{dx} = -sin[sin(2x^{3} - 2x + 4)[\cdot\dfrac{d[sin(2x^{3} - 2x + 4)]}{dx}}$

$:\implies \sf{\dfrac{dy}{dx} = -sin[sin(2x^{3} - 2x + 4)]\cdot[cos(2x^{3} - 2x + 4)\cdot(6x^{2} - 2)]}$

$\underline{\boxed{\therefore \sf{\dfrac{dy}{dx} = -sin[sin(2x^{3} - 2x + 4)]\cdot[cos(2x^{3} - 2x + 4)\cdot(6x^{2} - 2)]}}}$

Answer 2

Solution :

$</p><p>Function \: of \begin{gathered}\sf{y = cos[sin(2x^{3} - 2x + 4)]} \\ \\ \end{gathered} </p><p>\\ y=cos[sin(2x </p><p>3</p><p> −2x+4)]\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[cos(sin(2x^{3} - 2x + 4))]}{dx}} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> = </p><p>dx</p><p>d[cos(sin(2x </p><p>3</p><p> −2x+4))]</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\textsf{Now by applying the chain rule of differentiation, we get :} \\ \\ \underline{\sf{\bigstar\:Chain\: rule\:of \: differentiation :-}} \\ \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}} \\ \\ \textsf{Taking here,} \\ \bullet\:\sf{The\:value\:of\:y = cos(sin(2x^{3} - 2x + 4))} \\ \bullet\:\sf{The\:value\:of\:u = sin(2x^{3} - 2x + 4)} \\ \\ \end{gathered} </p><p>Now by applying the chain rule of differentiation, we get :</p><p>★Chainruleofdifferentiation:−</p><p> </p><p> </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> = </p><p>du</p><p>dy</p><p> </p><p> ⋅ </p><p>dx</p><p>du</p><p> </p><p> </p><p>Taking here,</p><p>∙Thevalueofy=cos(sin(2x </p><p>3</p><p> −2x+4))</p><p>∙Thevalueofu=sin(2x </p><p>3</p><p> −2x+4)</p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[cos(sin(2x^{3} - 2x + 4))]}{d[sin(2x^{3} - 2x + 4)]}\cdot\dfrac{d[sin(2x^{3} - 2x + 4)]}{dx}} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> = </p><p>d[sin(2x </p><p>3</p><p> −2x+4)]</p><p>d[cos(sin(2x </p><p>3</p><p> −2x+4))]</p><p> </p><p> ⋅ </p><p>dx</p><p>d[sin(2x </p><p>3</p><p> −2x+4)]</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = -sin[sin(2x^{3} - 2x + 4)]\cdot\dfrac{d[sin(2x^{3} - 2x + 4)]}{dx}} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> =−sin[sin(2x </p><p>3</p><p> −2x+4)]⋅ </p><p>dx</p><p>d[sin(2x </p><p>3</p><p> −2x+4)]</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\sf{Now,\:by\:differentiating,\:sin(2x^{3} - 2x + 4), w.r.t x,\:we\:get :} \\ \\ \end{gathered} </p><p>Now,bydifferentiating,sin(2x </p><p>3</p><p> −2x+4),w.r.tx,weget:</p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[sin(2x^{3} - 2x + 4)]}{dx}} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> = </p><p>dx</p><p>d[sin(2x </p><p>3</p><p> −2x+4)]</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\textsf{Now by applying the chain rule of differentiation, we get :} \\ \\ \underline{\sf{\bigstar\:Chain\: rule\:of \: differentiation :-}} \\ \\ \\ :\implies \sf{\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}} \\ \\ \textsf{Taking here,} \\ \bullet\:\sf{The\:value\:of\:y = sin(2x^{3} - 2x + 4)} \\ \bullet\:\sf{The\:value\:of\:u = 2x^{3} - 2x + 4} \\ \\ \end{gathered} </p><p>Now by applying the chain rule of differentiation, we get :</p><p>★Chainruleofdifferentiation:−</p><p> </p><p> </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> = </p><p>du</p><p>dy</p><p> </p><p> ⋅ </p><p>dx</p><p>du</p><p> </p><p> </p><p>Taking here,</p><p>∙Thevalueofy=sin(2x </p><p>3</p><p> −2x+4)</p><p>∙Thevalueofu=2x </p><p>3</p><p> −2x+4</p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[sin(2x^{3} - 2x + 4)]}{dx}} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> = </p><p>dx</p><p>d[sin(2x </p><p>3</p><p> −2x+4)]</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = \dfrac{d[sin(2x^{3} - 2x + 4)]}{d(2x^{3} - 2x + 4)}\cdot\dfrac{d(2x^{3} - 2x + 4)}{dx}} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> = </p><p>d(2x </p><p>3</p><p> −2x+4)</p><p>d[sin(2x </p><p>3</p><p> −2x+4)]</p><p> </p><p> ⋅ </p><p>dx</p><p>d(2x </p><p>3</p><p> −2x+4)</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = cos(2x^{3} - 2x + 4)\cdot\dfrac{d(2x^{3} - 2x + 4)}{dx}} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> =cos(2x </p><p>3</p><p> −2x+4)⋅ </p><p>dx</p><p>d(2x </p><p>3</p><p> −2x+4)</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = cos(2x^{3} - 2x + 4)\cdot\bigg[\dfrac{d(2x^{3})}{dx} - \dfrac{d(2x)}{dx} + \dfrac{d(4)}{dx}\bigg]} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> =cos(2x </p><p>3</p><p> −2x+4)⋅[ </p><p>dx</p><p>d(2x </p><p>3</p><p> )</p><p> </p><p> − </p><p>dx</p><p>d(2x)</p><p> </p><p> + </p><p>dx</p><p>d(4)</p><p> </p><p> ]</p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = cos(2x^{3} - 2x + 4)\cdot[6x^{2} - 2 + 0]} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> =cos(2x </p><p>3</p><p> −2x+4)⋅[6x </p><p>2</p><p> −2+0]</p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = cos(2x^{3} - 2x + 4)\cdot(6x^{2} - 2)} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> =cos(2x </p><p>3</p><p> −2x+4)⋅(6x </p><p>2</p><p> −2)</p><p> </p><p> </p><p></p><p>\begin{gathered}\therefore \sf{\dfrac{dy}{dx} = cos(2x^{3} - 2x + 4)\cdot(6x^{2} - 2)} \\ \\ \end{gathered} </p><p>∴ </p><p>dx</p><p>dy</p><p> </p><p> =cos(2x </p><p>3</p><p> −2x+4)⋅(6x </p><p>2</p><p> −2)</p><p> </p><p> </p><p></p><p>\begin{gathered}\textsf{Now, by substituting the derivative of}\: \\ \sf sin(2x^{3} - 2x + 4)\:in\: the\: Equation, we \:get : \\ \\ \end{gathered} </p><p>Now, by substituting the derivative of</p><p>sin(2x </p><p>3</p><p> −2x+4)intheEquation,weget:</p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = -sin[sin(2x^{3} - 2x + 4)[\cdot\dfrac{d[sin(2x^{3} - 2x + 4)]}{dx}} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> =−sin[sin(2x </p><p>3</p><p> −2x+4)[⋅ </p><p>dx</p><p>d[sin(2x </p><p>3</p><p> −2x+4)]</p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}:\implies \sf{\dfrac{dy}{dx} = -sin[sin(2x^{3} - 2x + 4)]\cdot[cos(2x^{3} - 2x + 4)\cdot(6x^{2} - 2)]} \\ \\ \end{gathered} </p><p>:⟹ </p><p>dx</p><p>dy</p><p> </p><p> =−sin[sin(2x </p><p>3</p><p> −2x+4)]⋅[cos(2x </p><p>3</p><p> −2x+4)⋅(6x </p><p>2</p><p> −2)]</p><p> </p><p> </p><p></p><p>\begin{gathered}\underline{\boxed{\therefore \sf{\dfrac{dy}{dx} = -sin[sin(2x^{3} - 2x + 4)]\cdot[cos(2x^{3} - 2x + 4)\cdot(6x^{2} - 2)]}}} \\ \\ \end{gathered} </p><p>∴ </p><p>dx</p><p>dy</p><p> </p><p> =−sin[sin(2x </p><p>3</p><p> −2x+4)]⋅[cos(2x </p><p>3</p><p> −2x+4)⋅(6x </p><p>2</p><p> −2)]</p><p> </p><p> </p><p> </p><p> </p><p>$