Answer :
Answer:
[tex]\dfrac{\text{d}y}{\text{d}x} =\dfrac{80x(5x^2+1)^3}{(1-4(5x^2+1)^4)^2}[/tex]
Step-by-step explanation:
[tex]\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If $y=f(u)$ and $u=g(x)$ then:\\\\$\dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times\dfrac{\text{d}u}{\text{d}x}$\\\end{minipage}}[/tex]
Given:
[tex]y=\dfrac{2u}{1-4u}, \quad u=(5x^2+1)^4[/tex]
Differentiate the two parts separately.
Step 1
Differentiate y with respect to u using the quotient rule.
[tex]\boxed{\begin{minipage}{5.5 cm}\underline{Quotient Rule for Differentiation}\\\\$\dfrac{\text{d}}{\text{d}x}\left[\dfrac{\text{f}(x)}{\text{g}(x)}\right]=\dfrac{\text{g}(x)\text{f}\:'(x)-\text{f}(x)\text{g}'(x)}{\left(\text{g}(x)\right)^2}$\\\end{minipage}}[/tex]
[tex]\textsf{Given}: \quad y=\dfrac{2u}{1-4u}[/tex]
Therefore:
[tex]\text{f}(u)=2u \implies \text{f}\:'(u)=2[/tex]
[tex]\text{g}(u)=1-4u \implies \text{g}\:'(u)=-4[/tex]
[tex]\begin{aligned} \implies \dfrac{\text{d}y}{\text{d}u} & =\dfrac{(1-4u)(2)-2u(-4)}{(1-4u)^2}\\\\& =\dfrac{2-8u+8u}{(1-4u)^2}\\\\& =\dfrac{2}{(1-4u)^2}\end{aligned}[/tex]
Step 2
Differentiate u with respect to x using the chain rule.
[tex]u=(5x^2+1)^4, \:\: \textsf{so let }u=w^4 \textsf{ where }w=5x^2+1[/tex]
Differentiate the two parts separately:
[tex]u=w^4\implies \dfrac{\text{d}u}{\text{d}w}=4w^3=4(5x^2+1)^3[/tex]
[tex]w=5x^2+1\implies \dfrac{\text{d}w}{\text{d}x}=10x[/tex]
Put everything into the chain rule formula:
[tex]\begin{aligned}\implies \dfrac{\text{d}u}{\text{d}x} & =\dfrac{\text{d}u}{\text{d}w} \times \dfrac{\text{d}w}{\text{d}x}\\\\ & =4(5x^2+1)^3(10x)\\\\ &=40x(5x^2+1)^3\end{aligned}[/tex]
Step 3
Put the differentiated parts from steps 1 and 2 into the chain rule formula:
[tex]\begin{aligned}\implies \dfrac{\text{d}y}{\text{d}x} & =\dfrac{\text{d}y}{\text{d}u}\times\dfrac{\text{d}u}{\text{d}x}\\\\& = \dfrac{2}{(1-4u)^2} \times 40x(5x^2+1)^3\\\\& = \dfrac{80x(5x^2+1)^3}{(1-4u)^2} \end{aligned}[/tex]
[tex]\textsf{Finally, substitute back in }\:\:u=(5x^2+1)^4:[/tex] :
[tex]\implies \dfrac{\text{d}y}{\text{d}x} =\dfrac{80x(5x^2+1)^3}{(1-4(5x^2+1)^4)^2}[/tex]
Differentiation Rules
[tex]\boxed{\begin{minipage}{4.8 cm}\underline{Differentiating $ax^n$}\\\\If $y=ax^n$, then $\dfrac{\text{d}y}{\text{d}x}=nax^{n-1}$\\\end{minipage}}[/tex]
[tex]\boxed{\begin{minipage}{4cm}\underline{Differentiating a constant}\\\\If $y=a$, then $\dfrac{\text{d}y}{\text{d}x}=0$\\\end{minipage}}[/tex]
Learn more about the chain rule here:
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