gradient of y=4x²-6x-5 at x=1.5

Calculate Derivative: To find the gradient of the curve at a particular point, we need to calculate the derivative of the function with respect to x. The derivative of a function gives us the slope of the tangent line at any point on the curve.Apply Power Rule: The function given is y = 4x² - 6x - 5. Let's find the derivative of this function, which we'll denote as y'. To differentiate y = 4x² - 6x - 5, we apply the power rule which states that the derivative of x^n is n*x^(n-1).Evaluate Derivative at x=1.5: Differentiating each term separately: The derivative of 4x² is 8x (since 2*4x^(2-1) = 8x). The derivative of -6x is -6 (since the derivative of x is 1, and the constant multiple rule allows us to pull the constant out in front). The derivative of -5 is 0 (since the derivative of a constant is 0). So, y' = 8x - 6.Substitute x=1.5: Now we need to evaluate the derivative at x = 1.5 to find the gradient of the curve at that point. Substitute x = 1.5 into y' to get the gradient. y'(1.5) = 8(1.5) - 6.Calculate Gradient: Calculate the value of y'(1.5): y'(1.5) = 8 * 1.5 - 6 = 12 - 6 = 6.

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gradient of $y=4x^2-6x-5$ at $x=1.5$

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Q. gradient of

y=4x^2-6x-5

x=1.5

Calculate Derivative: To find the gradient of the curve at a particular point, we need to calculate the derivative of the function with respect to $x$ . The derivative of a function gives us the slope of the tangent line at any point on the curve.
Apply Power Rule: The function given is $y = 4x^2 - 6x - 5$ . Let's find the derivative of this function, which we'll denote as $y'$ . To differentiate $y = 4x^2 - 6x - 5$ , we apply the power rule which states that the derivative of $x^n$ is $n \cdot x^{n-1}$ .
Evaluate Derivative at $x=1.5$ : Differentiating each term separately: $\newline$ The derivative of $4x^2$ is $8x$ (since $2\cdot4x^{2-1} = 8x$ ). $\newline$ The derivative of $-6x$ is $-6$ (since the derivative of $x$ is $1$ , and the constant multiple rule allows us to pull the constant out in front). $\newline$ The derivative of $-5$ is $0$ (since the derivative of a constant is $0$ ). $\newline$ So, $4x^2$ $1$ .
Substitute $x=1.5$ : Now we need to evaluate the derivative at $x = 1.5$ to find the gradient of the curve at that point. $\newline$ Substitute $x = 1.5$ into $y'$ to get the gradient. $\newline$ $y'(1.5) = 8(1.5) - 6$ .
Calculate Gradient: Calculate the value of $y'(1.5)$ : $\newline$ $y'(1.5) = 8 \times 1.5 - 6 = 12 - 6 = 6$ .