Find the derivative of the given function.f(x)=(x3+2x)⋅log5xf(x)=(x^3+2x)\cdot\log_5xf(x)=(x3+2x)⋅log5x

x3+2xxln⁡5+log⁡5x⋅(3x2+2)\frac{$$x^3+2x$$}{x\ln5}+\log_5x\cdot\$$left(3x^2+2$$\right)xln5x3+2x+log5x⋅(3x2+2)

Find the derivative of the given function. f ( x ) = ( x 3 + 2 x ) ⋅ log 5 x f(x)=(x^3+2x)\cdot\log_5x f ( x ) = ( x 3 + 2 x ) ⋅ lo g 5 x

In this problem, we're asked to find the derivative of the function f of x equals x cubed plus 2x times log base 5 of x. Now here, we can clearly see that we have 2 functions multiplying each other. Because of that, we're going to have to apply the product rule. Now most of you likely already know the product rule, but if you have not yet learned the product rule, you're gonna wanna circle back to this problem after you've learned that rule. But for the rest of you, let's get into this. We wanna find our derivative here, f prime of x. Now remember our memory tool for the product rule tells us left d right plus right d left. So starting with that left hand function, that's x cubed plus 2x. We're multiplying that by the derivative of our right hand function. The the derivative of a log base 5 of x is going to be 1 over x times the natural log of 5. Then we're adding this together with a right d left, so that is log base 5 of x times the derivative of this x cubed plus 2 x. That's gonna give me 3x squared plus 2. So this is my full derivative here. I can go ahead and write this as a fraction. So this is x cubed plus 2x over x times the natural log of 5 and then plus here. I'm gonna go ahead and move this polynomial out front, 3 x squared plus 2 times log base 5 of x, and this is my final derivative here. Let me know if you have any questions. I'll see you in the next one.

Find the derivative of the given function.f(x)=(x3+2x)⋅log⁡5⁡xf(x)=$$(x^3+2x$$)\cdot\$$log_5$$⁡xf(x)=(x3+2x)⋅log5⁡x

x3+2xxln⁡5+log⁡5x⋅(3x2+2)\frac{$$x^3+2x$$}{x\ln5}+\log_5x\cdot\$$left(3x^2+2$$\right)xln5x3+2x+log5x⋅(3x2+2)

(3x2+2)xln⁡5\frac{$$(3x^2+2$$)}{x\ln5}xln5(3x2+2)

(3x2+2)⋅log⁡5⁡$$x+x2+2(3x^2+2$$)\cdot\$$log_5$$⁡$$x+x^2+2(3x2+2$$)⋅log5⁡x+x2+2

(3x2+2+x2ln⁡5+2ln⁡5)⋅log⁡5x\$$left(3x^2+2$$+\frac{$$x^2$$}{\ln5}+\frac{2}{\ln5}\right)\cdot\log_5x(3x2+2+ln5x2+ln52)⋅log5x

4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Business Calculus

4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

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Multiple Choice

Find the derivative of the given function.
$f(x)=(x^3+2x)$\cdot\]\log$_5x$

$\frac{(3x^2+2)}{x\ln5}$

$(3x^2+2)$\cdot\]\log$_5x+x^2+2$

$$\left$(3x^2+2+$\frac{x^2}{\ln5}$+$\frac{2}{\ln5}\[\right$)$\cdot\]\log$_5x$

$$\frac{x^3+2x}{x\ln5}$+$\log$_5x$\cdot\]\left$(3x^2+2$\right$)$

Verified step by step guidance

Step 1: Recognize that the given function is a product of two functions: f(x) = (x^3 + 2x) * log_5(x). To find the derivative, we will use the product rule, which states that if f(x) = u(x) * v(x), then f'(x) = u'(x) * v(x) + u(x) * v'(x).

Step 2: Identify u(x) = x^3 + 2x and v(x) = log_5(x). Start by finding the derivative of u(x). The derivative of u(x) = x^3 + 2x is u'(x) = 3x^2 + 2.

Step 3: Next, find the derivative of v(x) = log_5(x). Recall that the derivative of log_a(x) is 1 / (x * ln(a)), where ln(a) is the natural logarithm of the base. Therefore, v'(x) = 1 / (x * ln(5)).

Step 4: Apply the product rule: f'(x) = u'(x) * v(x) + u(x) * v'(x). Substitute u(x), u'(x), v(x), and v'(x) into this formula. This gives f'(x) = (3x^2 + 2) * log_5(x) + (x^3 + 2x) * (1 / (x * ln(5))).

Step 5: Simplify the expression. For the second term, (x^3 + 2x) * (1 / (x * ln(5))), factor out x from the numerator and denominator to simplify. Combine the terms to express the derivative in its simplest form.

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Struggling with Business Calculus?

Find the derivative of the given function.f(x)=(x3+2x)⋅log5xf(x)=(x^3+2x)\(\cdot\]\log\)_5xf(x)=(x3+2x)⋅log5x

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Derivatives of Exponential & Logarithmic Functions practice set

Find the derivative of the given function.
$f(x)=(x^3+2x)\(\cdot\]\log\)_5x$