Example Question - derivative of function
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Derivative of a Function with Simplification
The function \( h(x) \) is given as: \[ h(x) = \frac{4x^3 - 7x + 8}{x} \] Before we find the derivative, we can simplify the function by dividing each term in the numerator by \( x \). This gives us: \[ h(x) = 4x^2 - 7 + \frac{8}{x} \] Now, let's find the derivative of \( h(x) \) with respect to \( x \) using the power rule and the derivative of a constant. The power rule states that the derivative of \( x^n \) with respect to \( x \) is \( n \cdot x^{n-1} \), and the derivative of a constant is 0. For the term \( \frac{8}{x} \), which can be written as \( 8x^{-1} \), we also use the power rule. Let's differentiate each term: 1. The derivative of \( 4x^2 \) with respect to \( x \) is \( 2 \cdot 4x = 8x \). 2. The derivative of \( -7 \) (a constant) with respect to \( x \) is \( 0 \). 3. The derivative of \( \frac{8}{x} \) (which is \( 8x^{-1} \)) with respect to \( x \) is \( -1 \cdot 8x^{-1 - 1} = -8x^{-2} \). Putting it all together, the derivative of \( h(x) \) is: \[ h'(x) = 8x - 0 - \frac{8}{x^2} \] \[ h'(x) = 8x - \frac{8}{x^2} \] So, the derivative of the function \( h(x) \) is \( 8x - \frac{8}{x^2} \).
Derivative of a Function with a Rational Expression
To find the derivative of the function \( h(x) \) given by: \[ h(x) = \frac{4x^3 - 7x + 8}{x} \] you can simplify the function before differentiating it by dividing each term in the numerator by \( x \): \[ h(x) = \frac{4x^3}{x} - \frac{7x}{x} + \frac{8}{x} \] \[ h(x) = 4x^2 - 7 + \frac{8}{x} \] Now, we'll differentiate each term separately: - The derivative of \( 4x^2 \) with respect to \( x \) is \( 8x \). - The derivative of a constant, \( -7 \), is 0. - The derivative of \( \frac{8}{x} \), which is \( 8x^{-1} \), is \( -8x^{-2} \) or \( -\frac{8}{x^2} \). So, the derivative of \( h(x) \), which we denote as \( h'(x) \), is: \[ h'(x) = 8x - 0 - \frac{8}{x^2} \] \[ h'(x) = 8x - \frac{8}{x^2} \] This is the final form of the derivative of \( h(x) \).
Derivative of a Function with Respect to x using Quotient Rule and Chain Rule
The image shows a mathematical problem asking to find the derivative with respect to \(x\) of the function \(f(x,y)\), which is provided as: \[ \frac{d}{dx} \left( \frac{2xy}{\sqrt{x^2 + y^2}} \right) \] To find this derivative, we will use the quotient rule along with the chain rule because the function has both \(x\) and \(y\) variables, where \(y\) could be a function of \(x\). The quotient rule is: \[ \left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2} \] where \(u = 2xy\) and \(v = \sqrt{x^2 + y^2}\). The derivative of \(u = 2xy\) with respect to \(x\) is \(u' = 2y + 2xy'\) (using the product rule and assuming \(y\) is a function of \(x\)). The derivative of \(v = \sqrt{x^2 + y^2}\) with respect to \(x\) is \(v' = \frac{1}{2\sqrt{x^2 + y^2}} \cdot (2x + 2yy')\) (using the chain rule and assuming \(y\) is a function of \(x\)). Simplifying gives \(v' = \frac{x + yy'}{\sqrt{x^2 + y^2}}\). Now, apply the quotient rule: \[ \frac{d}{dx} \left( \frac{2xy}{\sqrt{x^2 + y^2}} \right) = \frac{(2y + 2xy')\sqrt{x^2 + y^2} - 2xy\left(\frac{x + yy'}{\sqrt{x^2 + y^2}}\right)}{x^2 + y^2} \] Simplify the expression: \[ = \frac{2y(x^2 + y^2) + 2xy'y\sqrt{x^2 + y^2} - 2x^2y - 2xy^2y'}{(x^2 + y^2)^{3/2}} \] Now, combine like terms and simplify further: \[ = \frac{2y^3 + 2xy'y\sqrt{x^2 + y^2} - 2xy^2y'}{(x^2 + y^2)^{3/2}} \] If \(y\) is not a function of \(x\), \(y'\) would simply be 0, and the expression would be further simplified. However, the derivative expressed as it stands takes into consideration that \(y\) may be a function of \(x\). If you have more specific information about \(y\), such as whether it is a constant or a function of \(x\), you can simplify the expression accordingly.
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