suppose ( y=sin (5 x^{2}+2 x - 5) ). we can write ( y=sin (u) ), where ( u=square ). the leibnitz notation…

suppose ( y=sin (5 x^{2}+2 x - 5) ). we can write ( y=sin (u) ), where ( u=square ). the leibnitz notation for the chain rule is ( \frac{d y}{d x}=\frac{d y}{d u} cdot \frac{d u}{d x} ). the factors are ( \frac{d y}{d u}=square ) (written as a function of ( u ) ) and ( \frac{d u}{d x}=square ). now substitute in the function of ( x ) for ( u ) to get ( \frac{d y}{d x}=square ) (written as a function of ( x ) ).

suppose ( y=sin (5 x^{2}+2 x - 5) ). we can write ( y=sin (u) ), where ( u=square ). the leibnitz notation for the chain rule is ( \frac{d y}{d x}=\frac{d y}{d u} cdot \frac{d u}{d x} ). the factors are ( \frac{d y}{d u}=square ) (written as a function of ( u ) ) and ( \frac{d u}{d x}=square ). now substitute in the function of ( x ) for ( u ) to get ( \frac{d y}{d x}=square ) (written as a function of ( x ) ).

Answer

Explanation:

Step1: Find (u)

Given (y = \sin(5x^{2}+2x - 5)), if (y=\sin(u)), then (u = 5x^{2}+2x - 5).

Step2: Find (\frac{dy}{du})

Since (y=\sin(u)), using the derivative formula (\frac{d}{du}(\sin(u))=\cos(u)), we have (\frac{dy}{du}=\cos(u)).

Step3: Find (\frac{du}{dx})

Since (u = 5x^{2}+2x - 5), using the power rule (\frac{d}{dx}(ax^{n})=nax^{n - 1}), we get (\frac{du}{dx}=\frac{d}{dx}(5x^{2}+2x - 5)=10x + 2).

Step4: Find (\frac{dy}{dx})

By the chain - rule (\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}). Substitute (\frac{dy}{du}=\cos(u)) and (\frac{du}{dx}=10x + 2) into the formula. Then replace (u) with (5x^{2}+2x - 5), so (\frac{dy}{dx}=\cos(5x^{2}+2x - 5)\cdot(10x + 2)).

Answer:

(u = 5x^{2}+2x - 5); (\frac{dy}{du}=\cos(u)); (\frac{du}{dx}=10x + 2); (\frac{dy}{dx}=(10x + 2)\cos(5x^{2}+2x - 5))