Question

Use the chain rule to find the derivatives of f(t)=(2t^3+1/3t^2+1)^2

Answer 1

Answer:

2(2t^3+1/3t^2+1).(6t^2 +2/3t)

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Answer 2

Concept

There can be functions nested within each other if the function depends on multiple variables. Multiply the  chain of smaller derivatives  to get the total derivative.

Given

We have been given function of t which is  $f(t)=(\frac{2t^3+1}{3t^2+1} )^2$

Find

We are asked to determine the derivative of the given function of t by using chain rule .

Solution

It is given that $f(t)=(\frac{2t^3+1}{3t^2+1} )^2$ .

Taking quantity $\frac{2t^3+1}{3t^2+1}$ as x then differentiating with respect to t , we get

$f'(t)=2 (\frac{2t^3+1}{3t^2+1} )$   ......(1)

Now we will multiply by the derivative of the term which we have considered x which is $\frac{2t^3+1}{3t^2+1}$ . Now derivation of this term is given by quotient rule which is given by

$\frac{d(f(x))}{dxg(x)} =\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}$

Using this rule differentiation of $\frac{2t^3+1}{3t^2+1}$ , we get

$\frac{6t^2(3t^2+1)-(2t^3+1)(6t)}{(3t^2+1)^2} \\\\\frac{18t^4+6t^2-12t^4-6t}{(3t^2+1)^2} \\\\\frac{6t^4+6t^2-6t}{(3t^2+1)^2}$

Multiply this term in equation (1) , we get

$f'(t)=2 (\frac{2t^3+1}{3t^2+1} )\times \frac{6t^4+6t^2-6t}{(3t^2+1)^2}$

You don't need to memorize the chain rule formula. Instead, we can simply apply the derivative formula (that refers to x) and  multiply the result by the derivative of the expression that replaces x.

Therefore, the derivative of f(t) is $f'(t)=2 (\frac{2t^3+1}{3t^2+1} )\times \frac{6t^4+6t^2-6t}{(3t^2+1)^2}$ .

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