Question

Fine the derivative of
$y = 4x - 8 \div 3x + 11 \\ y = {x}^{2} \div {x}^{3} + 1 \\ y = 1 + {x}^{2} - x \div 1 - {x}^{2} + x$
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Answer 1

• ANSWER IS GIVEN AT END IN EACH PART.

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Step-by-step explanation:

(i) y= $\frac{4x-8}{3x+11}$

Now, $\frac{dy}{dx} = \frac{(4x-8)'.(3x+11) - (3x+11)'.(4x-8)}{(3x+11)^{2}}$ (by applying dividing rule)

⇒$\frac{dy}{dx}= \frac{4(3x+11) - 3(4x-8)}{(3x+11)^{2} }$

⇒$\frac{dy}{dx}= \frac{12x+44-12x+24}{(3x+11)^{2} }$

⇒$\frac{dy}{dx}= \frac{64}{(3x+11)^{2} }$(∵ Both 12x will cancel eachother)

Hence, this is the answer.

(ii)y=$\frac{x^{2} }{x^{3}+1 }$

Now,$\frac{dy}{dx}=\frac{(x^{2} )'.(x^{3}+1)-x^{2}.(x^{3} +1)'}{(x^{3}+1) ^{2} }$(by applying dividing rule)

⇒$\frac{dy}{dx}=\frac{2x.(x^{3}+1 )-x^{2} .(3x^{2} )}{(x^{3}+1) ^{2} }$

⇒$\frac{dy}{dx}=\frac{2x^{4}+2x-3x^{4} }{(x^{3}+1) ^{2} }$

⇒$\frac{dy}{dx}=\frac{-x^{4} +2x}{(x^{3}+1) ^{2} }$

Hence, this is the answer.

(iii)y= $\frac{1+x^{2} -x}{1-x^{2} +x}$

Now, $\frac{dy}{dx} =\frac{(1+x^{2} -x)'.(1-x^{2} +x)-(1+x^{2} -x).(1-x^{2} +x)'}{(1-x^{2} +x)^{2} }$(by applying dividing rule)

⇒$\frac{dy}{dx} =\frac{2x-1.(1-x^{2} +x)-(1+x^{2} -x).(-2x+1)}{(1-x^{2} +x)^{2} }$

⇒$\frac{dy}{dx} =\frac{2x-2x^{3}+2x^{2} -1+x^{2} -x-[-2x+1-2x^{3}+x^{2} +2x^{2} -x] }{(1-x^{2} +x)^{2} }$

⇒$\frac{dy}{dx} =\frac{x-2x^{3}+3x^{2} -1-[-3x+1-2x^{3}+3x^{2}] }{(1-x^{2} +x)^{2} }$

⇒$\frac{dy}{dx} =\frac{x-2x^{3}+3x^{2} -1+3x-1+2x^{3}-3x^{2} }{(1-x^{2} +x)^{2} }$

⇒$\frac{dy}{dx} =\frac{4x-2}{(1-x^{2} +x)^{2} }$ (∵ both $2x^{3}$ and $3x^{2}$ will cancel out eachother)

Hence, this is the answer.

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