Question

How do I solve this using chain rule?​

Answer 1

Answer:

$\dfrac{\text{d}y}{\text{d}x} =\dfrac{80x(5x^2+1)^3}{(1-4(5x^2+1)^4)^2}$

Step-by-step explanation:

$\boxed{\begin{minipage}{5.4 cm}\underline{Chain Rule for Differentiation}\\\\If y=f(u) and u=g(x) then:\\\\ \dfrac{\text{d}y}{\text{d}x}=\dfrac{\text{d}y}{\text{d}u}\times\dfrac{\text{d}u}{\text{d}x} \\\end{minipage}}$

Given:

$y=\dfrac{2u}{1-4u}, \quad u=(5x^2+1)^4$

Differentiate the two parts separately.

Step 1

Differentiate y with respect to u using the quotient rule.

$\boxed{\begin{minipage}{5.5 cm}\underline{Quotient Rule for Differentiation}\\\\ \dfrac{\text{d}}{\text{d}x}\left[\dfrac{\text{f}(x)}{\text{g}(x)}\right]=\dfrac{\text{g}(x)\text{f}\:'(x)-\text{f}(x)\text{g}'(x)}{\left(\text{g}(x)\right)^2} \\\end{minipage}}$

$\textsf{Given}: \quad y=\dfrac{2u}{1-4u}$

Therefore:

$\text{f}(u)=2u \implies \text{f}\:'(u)=2$

$\text{g}(u)=1-4u \implies \text{g}\:'(u)=-4$

$\begin{aligned} \implies \dfrac{\text{d}y}{\text{d}u} & =\dfrac{(1-4u)(2)-2u(-4)}{(1-4u)^2}\\\\& =\dfrac{2-8u+8u}{(1-4u)^2}\\\\& =\dfrac{2}{(1-4u)^2}\end{aligned}$

Step 2

Differentiate u with respect to x using the chain rule.

$u=(5x^2+1)^4, \:\: \textsf{so let }u=w^4 \textsf{ where }w=5x^2+1$

Differentiate the two parts separately:

$u=w^4\implies \dfrac{\text{d}u}{\text{d}w}=4w^3=4(5x^2+1)^3$

$w=5x^2+1\implies \dfrac{\text{d}w}{\text{d}x}=10x$

Put everything into the chain rule formula:

$\begin{aligned}\implies \dfrac{\text{d}u}{\text{d}x} & =\dfrac{\text{d}u}{\text{d}w} \times \dfrac{\text{d}w}{\text{d}x}\\\\ & =4(5x^2+1)^3(10x)\\\\ &=40x(5x^2+1)^3\end{aligned}$

Step 3

Put the differentiated parts from steps 1 and 2 into the chain rule formula:

$\begin{aligned}\implies \dfrac{\text{d}y}{\text{d}x} & =\dfrac{\text{d}y}{\text{d}u}\times\dfrac{\text{d}u}{\text{d}x}\\\\& = \dfrac{2}{(1-4u)^2} \times 40x(5x^2+1)^3\\\\& = \dfrac{80x(5x^2+1)^3}{(1-4u)^2} \end{aligned}$

$\textsf{Finally, substitute back in }\:\:u=(5x^2+1)^4:$ :

$\implies \dfrac{\text{d}y}{\text{d}x} =\dfrac{80x(5x^2+1)^3}{(1-4(5x^2+1)^4)^2}$

Differentiation Rules

$\boxed{\begin{minipage}{4.8 cm}\underline{Differentiating ax^n }\\\\If y=ax^n , then \dfrac{\text{d}y}{\text{d}x}=nax^{n-1} \\\end{minipage}}$

$\boxed{\begin{minipage}{4cm}\underline{Differentiating a constant}\\\\If y=a , then \dfrac{\text{d}y}{\text{d}x}=0 \\\end{minipage}}$

Learn more about the chain rule here:

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