Question

What is the derivative of $f(t)=(t \times ln(t),3t^2-t)$?

Answer 1

Hint: To find the derivative of a multi-dimensional function, we need to find the derivative of separate parts of the function. Here, there are two components of the function, so we find the derivative of both of them one by one and then we simply combine those two and write the derivative of the function given.

Complete step by step solution:
We have $f(t)=(t \times ln(t),3t^2-t)$, take the first component and name it as $f_1$, i.e.:
$f_1=t\times ln(t)$
We need to find the derivative of this function component. Since this function is a product of two separate functions, we apply product rule on this component.
Product rule is given by the following formula:
If ‘a(x)’ and ‘b(x)’ are two functions of ‘x’, then the derivative of their product is given by the formula:
$(a(x)b(x))`=a(x)b`(x)+a`(x)b(x)$
Here, we have ‘t’ and ‘ln(t)’ as functions of ‘t’, so by product rule, the derivative of the first component turns out to be:
$f_1`(t)=(t)\left(\dfrac{1}{t}\right)+(ln(t))(1)$
$\implies f_1`(t)=1+ln(t)$
We have used the following two formulas here:
$\dfrac{d(ln(t))}{t}=\dfrac{1}{t}$
$\dfrac{d(t)}{t}=1$
Now, we find the derivative of the second component $f_2=3t^2-t$,
We use the formula:
$\dfrac{d(ax^2)}{x}=a\times 2\times x=2ax$, where ‘a’ is a constant.
In this case the constant ‘a’ is 3 and the dependent variable is ‘t’, so we obtain the following derivative:
$f_2`(t)=2\times3\times t-1$
$\implies f_2`(t)=6t-1$
Now we have found the derivative of both the components separately, we now combine these and obtain the derivative as follows:
$f`(t)=(1+ln(t),6t-1)$
Hence, the derivative is found.

Note: Use the formulae correctly, do not confuse between the formulae of the derivatives. And take care of any calculation mistakes that may occur. There two different functions given above; do not confuse them by considering them as single functions.