Question Video: Finding the First Partial Derivative in a Multivariable Function of Two Variables Involving Exponential Functions

Find the first partial derivative of the function 𝑓(π‘₯, 𝑦) = 𝑒^(𝑦)/(π‘₯ + 𝑦²) with respect to π‘₯.

02:13

Video Transcript

Find the first partial derivative of the function 𝑓 of π‘₯, 𝑦 equals 𝑒 to the power of 𝑦 over π‘₯ plus 𝑦 squared with respect to π‘₯.

We’ve been given a multivariable function. That’s a function in terms of more than one variable, that’s π‘₯ and 𝑦. And we’ve been asked to find its first partial derivative. We denote the first partial derivative of a function 𝑓 with respect to π‘₯ by using the πœ• symbol. And we pronounce it del 𝑓 by del π‘₯. Now, the partial derivative of a function is its derivative with respect to one of the variables with the other ones held constant. So, in this case, we’re going to treat the variable 𝑦 as if it’s a constant. What that means is 𝑒 to the power of 𝑦 and 𝑦 squared are also constants. So, we’re actually going to be differentiating with respect to π‘₯ an expression of the form π‘Ž over π‘₯ plus 𝑏.

By rewriting this as π‘Ž times π‘₯ plus 𝑏 to the power of negative one, we can see we can use a special case of the chain rule to find its derivative. That is the general power rule. With this rule, we multiply the entire term by the exponent and then reduce that exponent by one. So, we get negative one times π‘Ž times π‘₯ plus 𝑏 to the power of negative two. We then multiply all of that by the derivative of the inner function. In general forms, that’s the derivative of π‘₯ plus 𝑏. Well, the derivative of π‘₯ is one and the derivative of 𝑏, since it’s a constant, is zero.

So, this simply becomes negative π‘Ž times π‘₯ plus 𝑏 to the power of negative two or negative π‘Ž over π‘₯ plus 𝑏 squared. Comparing our function to this general form, and we see that we can let π‘Ž be equal to 𝑒 to the power of 𝑦 and 𝑏 be equal to 𝑦 squared. Remember, it’s absolutely fine to do this because we’re finding the first partial derivative. And so, we’re treating 𝑦 as if it is a constant. In our general form then, we replace π‘Ž with 𝑒 to the power of 𝑦 and 𝑏 with 𝑦 squared.

And this means the first partial derivative of our function 𝑓 of π‘₯, 𝑦 equals 𝑒 to the power of 𝑦 over π‘₯ plus 𝑦 squared with respect to π‘₯ is negative 𝑒 to the power of 𝑦 over π‘₯ plus 𝑦 squared squared.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.