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Chapter 15: Problem 1

Calculate the derivative of \(y=3 x^{2}+\mathrm{e}^{x}\) when \(x=0.5\).

Short Answer

Expert verified
Answer: The derivative of the function when \(x = 0.5\) is \(3 + e^{0.5}\).

Step by step solution

01

Differentiating the given function

Differentiate the given function with respect to \(x\). The derivatives of \(x^2\) and \(e^x\) are \(2x\) and \(e^x\), respectively. Using these, the derivative of the function is: $$ \frac{dy}{dx} = \frac{d}{dx} (3x^2 + e^x) $$ Applying the sum rule of differentiation, we differentiate each term: $$ \frac{dy}{dx} = 3 \frac{d}{dx} (x^2) + \frac{d}{dx} (e^x) $$ Next, substitute the derivatives calculated above: $$ \frac{dy}{dx} = 3(2x) + e^x $$ Which simplifies to: $$ \frac{dy}{dx} = 6x + e^x $$ Now the expression for the derivative is found. We need to evaluate this expression when \(x=0.5\).
02

Evaluating the derivative at \(x=0.5\)

Substitute the given value of \(x\) into the expression for the derivative: $$ \frac{dy}{dx}(0.5) = 6(0.5) + e^{0.5} $$ Simplify the expression by performing the operations: $$ \frac{dy}{dx}(0.5) = 3 + e^{0.5} $$ The derivative of the function \(y=3x^2 + e^x\) when \(x=0.5\) is \(3 + e^{0.5}\).

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Most popular questions from this chapter

Find the rate of change of the following functions: (a) \(\frac{3 t^{3}-t^{2}}{2 t}\) (b) \(\ln \sqrt{x}\) (c) \((t+2)(2 t-1)\) (d) \(\mathrm{e}^{3 v}\left(1-\mathrm{e}^{v}\right)\) (e) \(\sqrt{x}(\sqrt{x}-1)\)

If \(f\) is a function of \(x\), write down two ways in which the derivative can be written.

Verify that \(y=A \sin k x+B \cos k x\), where \(A, B\), \(k\) constants is a solution of $$ y^{\prime \prime}+k^{2} y=0 $$

The function \(y(x)\) is given by \(y(x)=x^{3}-3 x\). Calculate the intervals on which \(y\) is (a) increasing, (b) decreasing.

Calculate \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) where \(y\) is given by (a) \(3 x^{4}-2 x+\ln x\) (b) \(\sin 5 x-5 \cos x\) (c) \((x+1)^{2}\) (d) \(\mathrm{e}^{3 x}+2 \mathrm{e}^{-2 x}+1\) (e) \(5+5 x+\frac{5}{x}+5 \ln x\)
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