What is the partial derivative fx of the function f(x,y)=x^2*2^(x*y)?

2 Answers | Add Yours

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The function given has two variables x and y. We have to find the partial derivative f'x, which implies that y can be considered as a constant.

f(x,y) = x^2*2^(x*y)

f'x = [x^2]'*2^(x*y) + x^2*[2^(x*y)]'

=> f'x = 2x*2^(x*y) + x^2*2^(x*y)*ln 2*y

=> f'x = 2^(x*y)*x [ 2 + x*ln 2 * y]

The required partial derivative is 2^(x*y)*x [ 2 + x*y*ln 2]

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll calculate the partial derivative fx, differentiating the expression of f(x,y), with respect to x, assuming that y is a constant.

fx = df/dx = d (x^2*2^(x*y))/dx

Since it is a product, we'll apply the product rule:

d (x^2*2^(x*y))/dx = (x^2)'*2^(xy) + x^2*(2^(xy))'

d (x^2*2^(x*y))/dx= 2x*2^(xy) + x^2*2^(xy)*ln 2*(xy)'

We'll factorize by 2^(xy):

d (x^2*2^(x*y))/dx= 2x*2^(xy) + x^2*2^(xy)*ln 2*(y)

d (x^2*2^(x*y))/dx= 2^(xy)*(2x + y*x^2*ln2)

The partial derivative is: fx = 2^(xy)*(2x + y*x^2*ln2)

We’ve answered 318,989 questions. We can answer yours, too.

Ask a question