Video: Finding the First and Second Derivative of a Combination of Root Functions

Find the first and second order derivatives of the function 𝐺(π‘Ÿ) = 3βˆšπ‘Ÿ βˆ’ 5 multiplied by the fifth root of π‘Ÿ.

02:31

Video Transcript

Find the first and second order derivatives of the function 𝐺 π‘Ÿ equals three root π‘Ÿ minus five multiplied by the fifth root of π‘Ÿ.

Now, the first thing I’m gonna do is actually rewrite our function in index form. So this is gonna give me 𝐺 π‘Ÿ is equal to three π‘Ÿ to the power of a half minus five π‘Ÿ to the power of fifth. And we got that by actually using one of our exponent rules that tells us that the π‘Žth root of π‘₯ is equal to π‘₯ to the power of one over π‘Ž. Okay, so what do we do now? Well, the first stage is actually to find our first order derivative. So what we wanna do is differentiate our function 𝐺 π‘Ÿ.

Now, in order to actually differentiate our function, what we’re gonna use is our general rule for differentiation. And our general rule tells us that if we have a function that’s in the form π‘Žπ‘₯ to the power of 𝑏, then the derivative of this function is gonna be equal to π‘Žπ‘, so the coefficient multiplied by the exponent, then π‘₯ the power of 𝑏 minus one, so we subtract one from the exponent. So therefore, we can actually apply that to the function that we’re trying to differentiate.

So our first term is gonna be three over two π‘Ÿ to the power of negative a half. And that’s because we multiplied our coefficient three by our exponent a half which gave us three over two. And then we reduced the exponent by one. So it went from a half to negative a half. And then our second term is gonna be minus five over five π‘Ÿ to the power of negative four over five. So therefore, we can say that our first order derivative is gonna be equal to three over two π‘Ÿ to the power of negative a half minus π‘Ÿ to the power of negative four over five. And we’ve done that because we simplified it because we have five over five which is just one. Okay, great. So that’s our first stage complete.

So now, we actually move on to our second stage which is to find the second order derivative. And in order to find our second derivative, what we’re actually gonna do is differentiate our first order derivative. So we know that the second order derivative is equal to the derivative of our first order derivative. So therefore, what we’re gonna do is actually apply our differentiation rule again. So therefore, we’re gonna have our second order derivative is equal to three over two multiplied by negative a half, so that’s our coefficient multiplied by our exponent, then π‘Ÿ to the power of negative a half minus one. Then this is minus negative four over five π‘Ÿ to the power of negative four over five minus one.

So therefore, if we simplify this, we’re gonna get that the second order derivative of our function π‘Ÿ is equal to negative three over four π‘Ÿ to the power of negative three over two plus four over five π‘Ÿ to power of negative nine over five.

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