I could really use the help on this as soon as possible.
what is the slope of the line tangent to the curve (/x)^2 at point (4,16)
check your typing.
Your curve would have an equation of the form
y = .... or f(x) = ....
secondly, (/x^2) is missing its numerator.
its the squareroot of x squared
To find the slope of the line tangent to the curve at a given point, you need to find the derivative of the function that represents the curve. In this case, the given curve is y = (1/x)^2.
To find the derivative, you can apply the power rule for derivatives. The power rule states that if the function is in the form f(x) = x^n, then its derivative f'(x) is given by f'(x) = n*x^(n-1).
In this case, you need to find the derivative of y = (1/x)^2. Let's break it down step by step:
Step 1: Apply the power rule to find the derivative of (1/x)^2 with respect to x:
dy/dx = -2*(1/x)^(2-1) * (1/x^(1-1))
dy/dx = -2*(1/x) * (1/x^0)
dy/dx = -2*(1/x) * (1/1)
dy/dx = -2/x
Step 2: Now that we have the derivative dy/dx, we can find the slope of the tangent line at a specific point. In this case, the given point is (4, 16).
Step 3: Substitute x = 4 into dy/dx:
slope = dy/dx = -2/4
slope = -1/2
Therefore, the slope of the line tangent to the curve y = (1/x)^2 at the point (4, 16) is -1/2.