Soal Soal Persamaan Garis Singgung Pada Kurva Dan Turunan III

The equation of the tangent line at the point (3,2) on the graph of y = x^2 - 4x + 5 is y = 2x - 4. This can be determined by finding the derivative of the given function, which is 2x - 4. The derivative represents the slope of the tangent line at any given point on the graph. So, at the point (3,2), the slope of the tangent line is 2. Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the point on the line and m is the slope, we can substitute the values (3,2) and 2 into the equation to get y = 2x - 4.

The equation of the tangent line to the curve y = x^3 - 3x + 3 at the point (0,3) can be found by taking the derivative of the curve and evaluating it at x = 0 to find the slope of the tangent line. The derivative of y = x^3 - 3x + 3 is 3x^2 - 3. Evaluating this at x = 0 gives a slope of -3. Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the given point and m is the slope, we can substitute in the values x1 = 0, y1 = 3, and m = -3 to get the equation 3x + y - 3 = 0.

The equation of the tangent line to the curve y = x^2 - 4x + 3 at the point (1,0) is y = -2x + 2. This can be determined by finding the derivative of the function y = x^2 - 4x + 3, which is y' = 2x - 4. Substituting x = 1 into y' gives the slope of the tangent line at x = 1, which is -2. The equation of a line with slope -2 passing through the point (1,0) can be written as y = -2x + 2. Therefore, y = -2x + 2 is the equation of the tangent line at the point (1,0).

The given expression is a polynomial function in the form of y = x^5 - 3x + 10. To find its derivative, we need to differentiate each term with respect to x. The derivative of x^5 is 5x^4, the derivative of -3x is -3, and the derivative of 10 (a constant) is 0. Therefore, the derivative of the given function is 5x^4 - 3.

The correct answer is y = 10x - 8. This can be determined by finding the derivative of the given function, which is y' = 6x - 2. The derivative represents the slope of the tangent line at any given point on the curve. To find the equation of the tangent line at x = 2, substitute x = 2 into the derivative to get y' = 6(2) - 2 = 10. This gives the slope of the tangent line. Using the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is the point of tangency, we can plug in the values x1 = 2, y1 = 3(2)^2 - 2(2) + 4 = 14 to get the equation y - 14 = 10(x - 2). Simplifying this equation gives y = 10x - 8.

The given equation of the tangent line is y = 10x + 8. In order for this line to be a tangent to the curve y = x^2 + ax + 9 at the point with x-coordinate 1, the slope of the tangent line must be equal to the derivative of the curve at x = 1. Taking the derivative of the curve y = x^2 + ax + 9 with respect to x gives 2x + a. Plugging in x = 1, we get 2(1) + a = 2 + a. Therefore, the slope of the tangent line is 2 + a. Since the given slope of the tangent line is 10, we can set up the equation 2 + a = 10 and solve for a. Solving this equation gives a = 8.

The equation of the tangent line to the parabola y = x^2 + 1 at the point with x-coordinate 2 is y = 4x - 3. This can be determined by taking the derivative of the parabola equation with respect to x, which gives the slope of the tangent line at any given point. Then, plugging in x = 2 into the derivative equation gives the slope of the tangent line at x = 2. Finally, using the point-slope form of a linear equation, the equation of the tangent line can be determined using the slope and the given point.

The given question is asking for the derivative of the function f(x) = (2x - 1)^2 (x + 2). To find the derivative, we need to apply the product rule. The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by u'(x)v(x) + u(x)v'(x). In this case, u(x) = (2x - 1)^2 and v(x) = (x + 2). Taking the derivative of u(x) and v(x) gives us u'(x) = 4(2x - 1) and v'(x) = 1. Applying the product rule, we get f'(x) = u'(x)v(x) + u(x)v'(x) = 4(2x - 1)(x + 2) + (2x - 1)^2(1) = (2x - 1)(6x + 7). Therefore, the correct answer is (2x - 1)(6x + 7).

The derivative of a function f(x) is found by using the power rule and the chain rule. In this case, the function f(x) is (6x - 3)^4 * (2x - 1). Applying the power rule and the chain rule, we can find the derivative f'(x). Evaluating f'(1) means plugging in x=1 into the derivative function. After calculating the derivative and plugging in x=1, we get the answer of 810.

The equation of the tangent line to the curve y = x^2 - 2/x at the point (1, -1) is given by 4x - y - 5 = 0.

The equation of the tangent line to the curve at the point where the abscissa is 4 can be found by finding the derivative of the curve and substituting the x-coordinate into it. The derivative of y = x^2 - 4x is dy/dx = 2x - 4. Substituting x = 4 into the derivative gives dy/dx = 2(4) - 4 = 4. Therefore, the slope of the tangent line is 4. Using the point-slope form of a line, y - y1 = m(x - x1), where (x1, y1) is the point on the line, the equation of the tangent line becomes y - (4^2 - 4(4)) = 4(x - 4), simplifying which gives y - 16 = 4x - 16, or y - 4x + 16 = 0.