Differentiation of polynomials and more

In this video, we are going to look at some more rules for differentiation. So in this first slide, we’re going to look at differentiating trigonometric functions. An important definition that you have to know is that if y is sine kx, where k is some constant, then if we differentiate y with respect to x, it’s equal to k multiplied by the cosine of kx. So in our example, we’re going to differentiate y is 4 sine x plus x squared. So we’re first going to look at differentiating 4 sine x. Now, we can see if we’re writing the form sine kx, we’ve got sine x. So k is 1 in this case.

So when we differentiate this using the rule, we’ve got 4 lots of sine x, so we will have 4 lots of cosine x using the definition in green. And then, we next need to differentiate x squared, which as a reminder, if y is x to the power of n, then if we differentiate y with respect to x, that is n lots of x to the power of n subtract 1. So in our case, n is 2. So we would have n is 2 multiplied by x to the power of n subtract 1, would be 2 subtract 1 is one, so we would just have x in this case. And we’ve solved the first question.

So we’re going to look at another definition when we’re differentiating trigonometric functions now, which is if y is instead cosine of kx for some constant k, then if we differentiate y with respect to x, it is minus k sine of kx. It’s negative k lots of sine kx. So we’re now going to differentiate the following, which is a mixture of a cosine function and the sine function. So we’re first going to differentiate 5 cosine 6x, which using the rule, we see that k is 6. And so if we differentiate cosine 6x using the rule, that would be negative 6 sine 6x. But as you can see, we’re differentiating 5 lots of cosine 6x.

So the answer for the first part is 5 lots of negative 6 sine 6x. And then next, we have to differentiate 2 sine of negative 4x. And as a reminder, if y is sine of kx, then dy/dx is k lots of cosine kx. So in our case, k is negative 4. So when we differentiate this, we’re going to have plus 2 lots of negative 4 cosine negative 4x, because we’ve got 2 lots of sine negative 4x that we’re differentiating. So when we simplify this by multiplying 5 by negative 6 and 2 by negative 4, we get the dy/dx is negative 30 of sine 6x subtract 8 lots of cosine negative 4x.

We also need to know how to differentiate exponential functions. Which we can see if y is ne to the power of kx, then if we differentiate that, we get k lots of n multiplied by e to the power of kx, where k and n are constants. So in our example, we’re going to differentiate y is 2e to the power of negative 5x plus 5 sine of 3x. So if we first differentiate 2e to the power of negative 5x, we can see in our definition n is 2, and k is negative 5. So when we differentiate this, we will have k is negative 5 multiplied by 2e to the negative 5x, as in red there.

We’ve then got to differentiate 5 sine of 3x. And we can do this by using the rule that if y is sine of kx, then if we differentiate that, it’s k lots of cosine kx. In our case, k is 3. So we will have plus 5 lots of 3 is k cosine 3x. And if we simplify that by multiplying negative 5 by 2, and 5 by 3, we get the following as the answer.

So in our final example, we’re going to look at differentiating logarithmic functions. So for logarithmic functions, we have if y is the natural logarithm of kx, then if we differentiate that with respect to x, we just get 1/x. So it doesn’t matter what the constant k is in the natural logarithm. So we’re now going to find the derivative of y is 3 lots of the natural logarithm of x plus 2x cubed, and we’re then going to find the gradient to that curve when x is 0.5. So using our rule, if we differentiate first 3 lots of the natural logarithm of x, we get 3 lots of 1/x.

And we’re then going to differentiate 2x cubed, which using the rule, if y is x to the power of n, then dy/dx is nx to the power of n subtract 1, where in our case, n is 3. We get 2 lots of 3x to the power of 3 subtract 1, which when we simplify all of this, we would have 3/x plus 6x squared. So for part B, we’re going to find the gradient to the curve y is 3 lots of the natural logarithm of x plus 2x cubed when x is 0.5.

So by substituting x is 0.5 into the formula we’ve just derived for dy/dx of both, that would be 3/x is 3/0.5 plus 6 times x squared would be 6 times 0.5 squared, which when we type that in a calculator, we get 7.5 is the gradient to the curve.

As an extension, can you find the tangent to the curve y is 3 lots of the natural logarithm of x plus 2x cubed when x is 1? Discuss your answer in the comments below this video. Thank you for watching this video.