Derivatives

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AP Calculus BC › Derivatives

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Evaluate $\frac{d}{dx}(xe^{\sin x})$

$(x\cos x+1)e^{\sin x}$

CORRECT

$xe^{\sin x}$

$(x\sin x +1)e^{\sin x}$

$(xe^{\cos{x}}+1)x$

None of the other answers

Explanation

To evaluate this derivative, we use the Product Rule.

$\frac{d}{dx}(xe^{\sin x})$

$= (x)(\cos xe^{\sin x})+(1)(e^{\sin x})$ . Use the Product Rule. Keep in mind that the derivative of $e^{\sin x}$ involves the Chain Rule.

$=(x\cos x +1)e^{\sin x}$ . Factor out an $e^{\sin x}$ .

Evaluate $\frac{d}{dx}(xe^{\sin x})$

$(x\cos x+1)e^{\sin x}$

CORRECT

$xe^{\sin x}$

$(x\sin x +1)e^{\sin x}$

$(xe^{\cos{x}}+1)x$

None of the other answers

Explanation

To evaluate this derivative, we use the Product Rule.

$\frac{d}{dx}(xe^{\sin x})$

$= (x)(\cos xe^{\sin x})+(1)(e^{\sin x})$ . Use the Product Rule. Keep in mind that the derivative of $e^{\sin x}$ involves the Chain Rule.

$=(x\cos x +1)e^{\sin x}$ . Factor out an $e^{\sin x}$ .

What is the first derivative of the following function?

$f'(x)=-lnxsin(sinx)cosx+\frac{cos(sinx)}{x}$

CORRECT

$f'(x)=-lnxcos^{2}xcos(sinx)$

$f'(x)=lnxsin(sinx)cosx+\frac{cos(sinx)}{x}$

$\frac{-cos(sinx)}{x^{2}}$

$\frac{-sin(sinx)cosx)}{x}$

Explanation

We use the product rule to differentiate this function. Applying it looks like this:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}lnx(cos(sinx))+lnx\frac{\mathrm{d} }{\mathrm{d} x}cos(sinx)$

This simplifies to:

$\frac{cos(sinx)}{x}+\frac{\mathrm{d} }{\mathrm{d} x}cos(sinx)lnx$

We apply the chain rule to differentiate , which becomes . Plugging this into the above equation gives us:

$\frac{cos(sinx)}{x}-lnxsin(sinx)cosx$ or $-lnsin(sinx)cosx+\frac{cos(sinx)}{x}$

What is the first derivative of the following function?

$f'(x)=-lnxsin(sinx)cosx+\frac{cos(sinx)}{x}$

CORRECT

$f'(x)=-lnxcos^{2}xcos(sinx)$

$f'(x)=lnxsin(sinx)cosx+\frac{cos(sinx)}{x}$

$\frac{-cos(sinx)}{x^{2}}$

$\frac{-sin(sinx)cosx)}{x}$

Explanation

We use the product rule to differentiate this function. Applying it looks like this:

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}lnx(cos(sinx))+lnx\frac{\mathrm{d} }{\mathrm{d} x}cos(sinx)$

This simplifies to:

$\frac{cos(sinx)}{x}+\frac{\mathrm{d} }{\mathrm{d} x}cos(sinx)lnx$

We apply the chain rule to differentiate , which becomes . Plugging this into the above equation gives us:

$\frac{cos(sinx)}{x}-lnxsin(sinx)cosx$ or $-lnsin(sinx)cosx+\frac{cos(sinx)}{x}$

Evaluate the derivative of , where is any constant.

CORRECT

None of the other answers

Explanation

For the term, we simply use the power rule to abtain . Since is a constant (not a variable), we treat it as such. The derivative of any constant (or "stand-alone number") is .

Evaluate the derivative of , where is any constant.

CORRECT

None of the other answers

Explanation

For the term, we simply use the power rule to abtain . Since is a constant (not a variable), we treat it as such. The derivative of any constant (or "stand-alone number") is .

What is the derivative of

$f(x)=5\sqrt{x}$ ?