Video Transcript
Find the derivative of the function
𝑦 equals five 𝑥 squared 𝑒 to the power of one over 𝑥.
So the first thing we can do, if
we’re gonna try and differentiate our function, is take out the constant term
because this won’t affects the differentiation. So, we’ve got the five. So, we can rewrite our function as
the derivative of 𝑦 is equal to five, because as we said, we’ve taken the constant
term outside, multiplied by the derivative of 𝑥 squared multiplied by 𝑒 to the
power of one over 𝑥.
So now, what we need to do in order
that we can differentiate 𝑥 squared multiplied by 𝑒 to the power of one over 𝑥 is
use the product rule. Now, the product rule tells us is
that if we have 𝑓 of 𝑥 multiplied by 𝑔 of 𝑥, that we can call 𝑓 of 𝑥 𝑢, 𝑔 of
𝑥 𝑣, and then we can say the derivative of 𝑢𝑣 is gonna be equal to 𝑢d𝑣 d𝑥
plus 𝑣d𝑢 d𝑥. So, it’s one of our terms
multiplied by the derivative of the other term plus the other term multiplied by the
derivative of our first term. Okay so, now we have this product
rule. Let’s apply it.
So, what we’re gonna do is call our
𝑢 𝑥 squared and our 𝑣 𝑒 to the power of one over 𝑥. And therefore, if we differentiate
𝑥 squared, so we find d𝑢 d𝑥, this is gonna be two 𝑥. And that’s because what we do is we
multiply the exponent by the coefficient, so in this case two multiplied by one. And then, what we do is we reduce
the exponent by one. So, it goes from two to one, so
we’re left with two 𝑥.
And now, what we need to do is
differentiate 𝑒 to the power of one over 𝑥. And in order to do that, what we’re
gonna have to do is use an exponential function rule for a differentiation. And that tells us if we want to
find the derivative of 𝑒 to the power of 𝑢 of 𝑥, then this is equal to 𝑒 to the
power of 𝑢 of 𝑥 multiplied by the derivative of 𝑢 of 𝑥. So therefore, in our case, it’s
gonna be d𝑣 d𝑥 is equal to 𝑒 to the power of one over 𝑥 multiplied by the
derivative of one over 𝑥.
So, to differentiate one over 𝑥,
what we’re gonna do is use the reciprocal rule. And, this tells us if we want the
derivative of one over 𝑢 of 𝑥, this is equal to negative and then the derivative
of 𝑢 of 𝑥 over 𝑢 of 𝑥 squared. So, what this is gonna give us is
𝑒 to the power of one over 𝑥, that’s what we already had. And then for the derivative of one
over 𝑥, we’re gonna have negative then we’ve got one as the numerator. And that’s because the derivative
of 𝑥 is just one and then over 𝑥 squared, because that’s our 𝑢 of 𝑥 squared.
So therefore, we’re gonna get
negative 𝑒 to the power of one over 𝑥 over 𝑥 squared. So then, what we do is we apply our
product rule. So, we’re gonna get the derivative
of 𝑥 squared multiplied by 𝑒 to the power of one over 𝑥 is equal to. Then, we’ve got our 𝑢d𝑣 d𝑥. So, 𝑥 squared multiplied by
negative 𝑒 to the power of one over 𝑥 over 𝑥 squared plus then we’ve got our
𝑣d𝑢 d𝑥. So, it’s 𝑒 to the power of one
over 𝑥 multiplied by two 𝑥.
So, we can simplify. So, we’re gonna see that our 𝑥
squareds are going to cancel on the left-hand term. That’s cause we got 𝑥 squared
multiplied by something over 𝑥 squared. So therefore, we’re gonna be left
with negative 𝑒 to the power of one over 𝑥 plus two 𝑥. Multiplied by 𝑒 to the power of
one over 𝑥.
So now, if we put that back into
our original expression, we’ve got the derivative of 𝑦 is equal to five. Multiplied by then we’ve got two
𝑥𝑒 to the power of one over 𝑥 minus 𝑒 to the power of one over 𝑥.
But have we finished? Well, no. Because we can simplify a step
further. Because what we can do is see that
each of our terms has a common factor. And that factor is 𝑒 to the power
of one over 𝑥. So therefore, we can take this
outside our parentheses. So therefore, what we can say is
that the derivative of the function 𝑦 equals five 𝑥 squared multiplied by 𝑒 to
the power of one over 𝑥 is five 𝑒 to the power of one over 𝑥 multiplied by two 𝑥
minus one.
