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

25x[5ln⁡2(4x−3x2+9)+(4−6x)]$$2^{5x}$$\left\lbrack5\ln2\$$left(4x-3x^2+9$$\right)+\left(4-6x\right)\right\rbrack

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[5ln⁡2(4x−3x2+9)+(4−6x)]$$2^{5x}$$\left\lbrack5\ln2\$$left(4x-3x^2+9$$\right)+\left(4-6x\right)\right\rbrack

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

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Calculus

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Calculus 때문에 힘들어하고 계신가요?

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객관식

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

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

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

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

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

검증된 단계별 안내

Identify the function as a product of two functions: u(x) = (4x - 3x^2 + 9) and v(x) = 2^{5x}.

Apply the product rule for differentiation, which states that the derivative of a product u(x)v(x) is u'(x)v(x) + u(x)v'(x).

Differentiate u(x) = 4x - 3x^2 + 9 to get u'(x) = 4 - 6x.

Differentiate v(x) = 2^{5x} using the chain rule. The derivative is v'(x) = 2^{5x} * 5ln(2).

Substitute u(x), u'(x), v(x), and v'(x) into the product rule formula to find the derivative: y' = (4 - 6x)2^{5x} + (4x - 3x^2 + 9) * 2^{5x} * 5ln(2).

6:32m

다음 시청

마스터 Derivatives of General Logarithmic Functions가 Patrick Ford 짧은 영상으로 설명해 줍니다

Derivatives of Exponential & Logarithmic Functions 연습 세트

수석 튜터가 만든 문제 세트

전문가 영상 설명

Calculus 때문에 힘들어하고 계신가요?

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

다음 시청

Derivatives of Exponential & Logarithmic Functions 연습 세트

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