If 𝑓 of 𝑥 is equal to three ln
multiplied by two 𝑥 plus four ln 𝑥, find the derivative of 𝑓 of 𝑥 when 𝑥 is
equal to one.
Now, the first thing we can do if
we’re looking to differentiate our function is take our constant term out because
this wouldn’t affect the differentiation. So we’ve now got three multiplied
by and then the derivative of ln multiplied by two 𝑥 plus four ln 𝑥. Now in order to solve the problem
and work out the derivative of the expression, we’re gonna have to use one of our
derivative rules. And this is one that’s based around
our natural logarithm, which is ln.
So we know that if we want to find
the derivative of ln — and in this case, we’re gonna say 𝑢 of 𝑥, so just a
function — then it’s equal to one over 𝑢 of 𝑥 multiplied by the derivative of 𝑢
of 𝑥. So therefore, what we’re gonna get
is three multiplied by and then we’ve got one over two 𝑥 plus four ln 𝑥. And that’s because that was our 𝑢
of 𝑥. So that was our function when we’re
looking at the rule for differentiating something like this. Then, we multiply this by the
derivative of two 𝑥 plus four ln 𝑥.
And as we know with
differentiation, what we can do is when we’re differentiating an expression like
this is we can differentiate each term separately. So we can differentiate two 𝑥 and
we can differentiate four ln 𝑥. So first of all, if we
differentiate two 𝑥, we’re just gonna get two. And that’s because what we do is we
multiply the coefficient by the exponent. So we’d have two multiplied by one
which is just two. And then, we reduce exponent from
one to zero. So we get two multiplied by one
which is just two.
And then, what we’re gonna do to be
able to differentiate the second term is use another one of our rules. And that is if we have derivative
of ln 𝑥, it’s just equal to one over 𝑥. So if we’re looking to
differentiate for ln 𝑥, the first thing we can do — as normal — is take the four,
the constant, out. So we have four multiplied by the
derivative of ln 𝑥, which is just gonna give us four over 𝑥. And that’s because we’re gonna have
four multiplied by one over 𝑥 which is four over 𝑥. So what this does is that it leaves
us with an expression. And this expression is three
multiplied by two because that was the derivative of two 𝑥 plus four over 𝑥
because that was the derivative of four ln 𝑥. And this is all over two 𝑥 plus
four ln 𝑥.
Now, at this stage, we’d think
about maybe simplifying. But there’s no need because what
we’re trying to find is the value of this expression, so the value of the
derivative, when 𝑥 is equal to one. And therefore, to do this, what we
need to do is we need to substitute in one for 𝑥 at every point in our derivative
expression. So when we do that, we get three
multiplied by two plus four over one over two multiplied by one plus four ln
one. So then, what we’re gonna get is 18
over two. And that’s because we got three
multiplied by two plus four. Well, two plus four is six. Three sixes are 18. Then I’ll need denominator: we’ve
got two multiplied by one which is two then add four ln one, well ln one is just
equal to zero. So we’re left with two on the
denominator. So therefore, this is gonna give us
a final answer of nine.
So therefore, we can say that if
function 𝑓 of 𝑥 is equal to three ln two 𝑥 plus four ln 𝑥, then the first
derivative value when 𝑥 is equal to one is going to be nine.