Find the derivative of the given function.y=(4x−3x2+9)⋅25xy=(4x-3x^2+9)\cdot2^{5x}y=(4x−3x2+9)⋅25x

25x⋅[(4−6x)+5ln⁡⁡(2)(4x−3x2+9)]$$2^{5x}$$\cdot[(4-6x)+5\ln⁡$$(2)(4x-3x^2+9$$)]25x⋅[(4−6x)+5ln⁡(2)(4x−3x2+9)]

Find the derivative of the given function. y = ( 4 x − 3 x 2 + 9 ) ⋅ 2 5 x y=(4x-3x^2+9)\cdot2^{5x} y = ( 4 x − 3 x 2 + 9 ) ⋅ 2 5 x

Here, we wanna find the derivative of the function y equals 4x minus 3x squared plus 9 times 2 to the power of 5x. Now at this point, if you have not yet learned your product rule and your chain rule, you will not be able to solve this problem. So go ahead and circle back to this after you've gone through your other rules for finding derivatives. If you've already learned all of your rules for finding derivatives, then let's go ahead and jump in. Here, we have 2 functions being multiplied. We have 4 x minus 3 x squared plus 9 times 2 to the power of 5 x. So that means we need to apply our product rule here. Now remember that our rule or our memory tool for the product rule tells us a left d right plus a right d left, where d just means we're taking it, the derivative. So I'm gonna start with that left hand function, leaving it as is. That's 4x minus 3x squared plus 9. Then I'm multiplying that by the derivative of that right hand function. So this is left d right. Now looking at this, this is an exponential function. But because that power is not just x, it is 5x, that means I also need to apply the chain rule here. So going ahead and taking the derivative of my outside function, that's what we do using the chain rule. That's 2 to the power of 5 x times the natural log of 2. Then applying the chain rule, I am multiplying them by the derivative of my inside function, which in this case is 5 x. The derivative of 5 x is just 5. So that term is left d right. We're adding this together with a right d left. So that is 2 to the power of 5 x, leaving that function as is, multiplied by the derivative of this polynomial function. Now we've taken the derivative of a bunch of a bunch of polynomial functions. So we know that this is going to be just 4. Then applying the power rule here, that is minus 6x. And we're all good here. Now I can do a little bit of cleaning up here because both of these terms have that 2 to the power of 5x in common. I can go ahead and factor that out. So this is 2 to the power of 5x. And then inside of my parentheses here, I'm going to go ahead and move that 5 times the natural log of 2 out front. It's multiplied by that polynomial, 4x minus 3x squared plus 9. And then this is added together with that 4 minuteus 6x, making sure I'm putting all of my parentheses where needed. And that is my full derivative, dydx. Let me know if you have any questions here, and let's continue practicing.

Find the derivative of the given function.y=(4x−3x2+9)⋅25xy=$$(4x-3x^2+9$$)\$$cdot2^{5x}y=(4x$$−3x2+9)⋅25x

25x⋅[(4−6x)+5ln⁡⁡(2)(4x−3x2+9)]$$2^{5x}$$\cdot[(4-6x)+5\ln⁡$$(2)(4x-3x^2+9$$)]25x⋅[(4−6x)+5ln⁡(2)(4x−3x2+9)]

25x⋅(−15x2−$$14x+49)2^{5x}$$\$$cdot(-15x^2-14x+49)25x$$⋅(−15x2−14x+49)

25x[(4−6x)+ln⁡⁡(2)(4x−3x2+9)]$$2^{5x}$$[(4-6x)+\ln⁡$$(2)(4x-3x^2+9$$)]25x[(4−6x)+ln⁡(2)(4x−3x2+9)]

5ln⁡⁡(2)(4−6x)⋅25x5\ln⁡(2)(4-6x)\$$cdot2^{5x}5ln$$⁡(2)(4−6x)⋅25x

4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Business Calculus

4. Derivatives of Exponential & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Struggling with Business Calculus?

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Multiple Choice

Find the derivative of the given function.
$y=(4x-3x^2+9)$\cdot$2^{5x}$

$2^{5x}$\cdot$[(4-6x)+5$\ln$(2)(4x-3x^2+9)]$

$2^{5x}$\cdot$(-15x^2-14x+49)$

$2^{5x}[(4-6x)+$\ln$(2)(4x-3x^2+9)]$

$5$\ln$(2)(4-6x)$\cdot$2^{5x}$

Verified step by step guidance

Step 1: Recognize that the given function y = (4x - 3x^2 + 9) * 2^(5x) is a product of two functions: u(x) = (4x - 3x^2 + 9) and v(x) = 2^(5x). To find the derivative, apply the product rule: (uv)' = u'v + uv'.

Step 2: Differentiate u(x) = (4x - 3x^2 + 9). Use the power rule for each term: the derivative of 4x is 4, the derivative of -3x^2 is -6x, and the derivative of 9 is 0. Thus, u'(x) = 4 - 6x.

Step 3: Differentiate v(x) = 2^(5x). Recall that the derivative of an exponential function a^(g(x)) is a^(g(x)) * ln(a) * g'(x). Here, a = 2 and g(x) = 5x. The derivative is v'(x) = 2^(5x) * ln(2) * 5.

Step 4: Substitute u(x), u'(x), v(x), and v'(x) into the product rule formula. This gives: y' = u'(x)v(x) + u(x)v'(x). Substitute u'(x) = (4 - 6x), v(x) = 2^(5x), u(x) = (4x - 3x^2 + 9), and v'(x) = 2^(5x) * ln(2) * 5.

Step 5: Simplify the expression for y'. Combine like terms and factor where possible to express the derivative in its simplest form. The final expression will involve terms with 2^(5x), (4 - 6x), and ln(2)(4x - 3x^2 + 9).

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Struggling with Business Calculus?

Find the derivative of the given function.y=(4x−3x2+9)⋅25xy=(4x-3x^2+9)\(\cdot\)2^{5x}y=(4x−3x2+9)⋅25x

Watch next

Derivatives of Exponential & Logarithmic Functions practice set

Find the derivative of the given function.
$y=(4x-3x^2+9)\(\cdot\)2^{5x}$