Question

If u=sin(ax+by+cz)then find

Answer 1

We recall the concept of partial differentiation

Partial differentiation is the method of finding partial differentiation

Given:

$u=sin(ax+by+cz)$

Partial differentiation is given below

$\frac{du}{dx} =acos(ax+by+cz)$

$\frac{du}{dy} =bcos(ax+by+cz)$

$\frac{du}{dz} =ccos(ax+by+cz)$

Answer 2

Answer:

$\frac{dy}{dx}=acos(ax+by+cz)\\ \frac{dy}{dx} =bcos(ax+by+cz)\\\frac{dy}{dx}=ccos(ax+by+cz)$

Step-by-step explanation:

Given:

u=sin(ax+by+cz)

Partial differentiation is given below:

$\frac{dy}{dx}=acos(ax+by+cz)\\ \frac{dy}{dx} =bcos(ax+by+cz)\\\frac{dy}{dx}=ccos(ax+by+cz)$

The derivative of a function of one variable tells us how fast the value of the function changes as the value of the independent variable changes. Intuitively, it tells us how "steep" the graph of a function is. We might wonder if there is a similar idea for graphs of functions of two variables, i.e. surfaces. It is not clear that this has a simple answer, nor how we might proceed. We start with what seems to be very small steps towards the goal. Surprisingly, these simple ideas turn out to be the key to a more general understanding.

The partial derivative of a function (in two or more variables) is its derivative with respect to one of the variables, all other variables being constant. The process of calculating the partial derivative is the same as the ordinary derivative, except that we consider variables other than the variable we are deriving as constants.

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