$f(x,y)=ln(4x-3y)$, (5,-7)

So, to do this I just find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ right?

If so, then:

$\frac{\partial f}{\partial x}=\frac{4}{4x-3y}$

$\frac{\partial f}{\partial y}=\frac{-3}{4x-3y}$

And then:

$f(5,-7)=\frac{4}{4(5)-3(-7)}i-\frac{3}{4(5)-3(-7)}j=>\frac{4}{41}i-\frac{3}{41}j$

Is this correct? Thanks.

2. Originally Posted by downthesun01
$f(x,y)=ln(4x-3y)$, (5,-7)

So, to do this I just find $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ right?

If so, then:

$\frac{\partial f}{\partial x}=\frac{4}{4x-3y}$

$\frac{\partial f}{\partial y}=\frac{-3}{4x-3y}$

And then:

$f(5,-7)=\frac{4}{4(5)-3(-7)}i-\frac{3}{4(5)-3(-7)}j=>\frac{4}{41}i-\frac{3}{41}j$

Is this correct? Thanks.
The question has no answer because there is no unique gradient at the point. Go back and check the question. It will no doubt ask for the gradient in a particular direction at the given point. The answer will be given by the directional derivative at the given point.

3. There may be a "British English", "American English" confusion here. In America we speak of things like df/dx as the "derivative", not gradient. The gradient of a function of two variables is specifically the vector $\nabla f$ which is exactly what downthesun01 has.

We would then talk about the "derivative" in a particular direction, say at angle $\theta$ to the positive x-axis, as $cos(\theta)\frac{\partial f}{\partial x}+ sin(\theta)\frac{\partial f}{\partial y}$, which is the same as the dot product of $\nabla f$ with a unit vector in that direction, which is what mr. fantastic appears to be calling the "gradient".

4. Originally Posted by HallsofIvy
There may be a "British English", "American English" confusion here. In America we speak of things like df/dx as the "derivative", not gradient. The gradient of a function of two variables is specifically the vector $\nabla f$ which is exactly what downthesun01 has.

We would then talk about the "derivative" in a particular direction, say at angle $\theta$ to the positive x-axis, as $cos(\theta)\frac{\partial f}{\partial x}+ sin(\theta)\frac{\partial f}{\partial y}$, which is the same as the dot product of $\nabla f$ with a unit vector in that direction, which is what mr. fantastic appears to be calling the "gradient".
The simple explanation is that I misunderstood the question. The OP appears to have the correct answer.