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$Differentiate these using the product rule. Write your answers in fully factorised form with the common factor before the brackets. {:[y=x^(5)(x^(2)+3)],[y=e^(x)(x^(5)+1)],[(dy)/(dx)=],[" [2] "(dy)/(dx)=e^(◻)" ( "],[y=x^(6)(ln x+1)],[(dy)/(dx)=],[" [2] "(dy)/(dx)=e],[y=e^(2x)(5x+2)],[y=x^(5)e^(2x)],[y=5x(x^(2)-2)^(3)],[(dy)/(dx)=◻e^(◻)(◻)],[" [3] "(dy)/(dx)=]:}$

Differentiate these using the product rule. $\newline$ Write your answers in fully factorised form with the common factor before the brackets. $\newline$ $\begin{array}{l} y=x^{5}\left(x^{2}+3\right) \\ y=e^{x}\left(x^{5}+1\right) \\ \frac{d y}{d x}= \\ \text { [2] } \frac{d y}{d x}=e^{\square} \text { ( } \\ y=x^{6}(\ln x+1) \\ \frac{d y}{d x}= \\ \text { [2] } \frac{d y}{d x}=e \\ y=e^{2 x}(5 x+2) \\ y=x^{5} e^{2 x} \\ y=5 x\left(x^{2}-2\right)^{3} \\ \frac{d y}{d x}=\square e^{\square}(\square) \\ \text { [3] } \frac{d y}{d x}= \\ \end{array}$

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Math Problems

Algebra 1

Compare linear and exponential growth

Full solution

Q. Differentiate these using the product rule.

\newline

Write your answers in fully factorised form with the common factor before the brackets.

\newline

\begin{array}{l} y=x^{5}\left(x^{2}+3\right) \\ y=e^{x}\left(x^{5}+1\right) \\ \frac{d y}{d x}= \\ \text { [2] } \frac{d y}{d x}=e^{\square} \text { ( } \\ y=x^{6}(\ln x+1) \\ \frac{d y}{d x}= \\ \text { [2] } \frac{d y}{d x}=e \\ y=e^{2 x}(5 x+2) \\ y=x^{5} e^{2 x} \\ y=5 x\left(x^{2}-2\right)^{3} \\ \frac{d y}{d x}=\square e^{\square}(\square) \\ \text { [3] } \frac{d y}{d x}= \\ \end{array}

Apply Product Rule: To differentiate the first function $y = x^5(x^2 + 3)$ , we apply the product rule which states that $(fg)' = f'g + fg'$ , where $f = x^5$ and $g = x^2 + 3$ .
Differentiate $f$ : Differentiate $f = x^5$ to get $f' = 5x^4$ .
Differentiate $g$ : Differentiate $g = x^2 + 3$ to get $g' = 2x$ .
Apply Product Rule: Now apply the product rule: $(\frac{dy}{dx}) = f'g + fg' = 5x^4(x^2 + 3) + x^5(2x)$ .
Simplify Expression: Simplify the expression: $(\frac{dy}{dx}) = 5x^6 + 15x^4 + 2x^6$ .
Combine Like Terms: Combine like terms: $(\frac{dy}{dx}) = 7x^6 + 15x^4$ .
Factor Out Common Factor: Factor out the common factor: $(\frac{dy}{dx}) = x^4(7x^2 + 15)$ .
Apply Product Rule: For the second function $y = e^x(x^5 + 1)$ , let $f = e^x$ and $g = x^5 + 1$ .
Simplify Expression: Differentiate $f = e^x$ to get $f' = e^x$ .
Factor Out Common Factor: Differentiate $g = x^5 + 1$ to get $g' = 5x^4$ .
Apply Product Rule: Apply the product rule: $(\frac{dy}{dx}) = f'g + fg' = e^x(x^5 + 1) + e^x(5x^4)$ .
Differentiate $f$ : Simplify the expression: $\frac{dy}{dx} = e^x(x^5 + 1 + 5x^4)$ .
Differentiate $g$ : Factor out the common factor: $\frac{dy}{dx} = e^x(x^5 + 5x^4 + 1)$ .
Apply Product Rule: For the third function $y = x^6(\ln x + 1)$ , let $f = x^6$ and $g = \ln x + 1$ .
Simplify Expression: Differentiate $f = x^6$ to get $f' = 6x^5$ .
Factor Out Common Factor: Differentiate $g = \ln x + 1$ to get $g' = \frac{1}{x}$ .
Apply Product Rule: Apply the product rule: $(\frac{dy}{dx}) = f'g + fg' = 6x^5(\ln x + 1) + x^6(\frac{1}{x})$ .
Simplify Expression: Simplify the expression: $(\frac{dy}{dx}) = 6x^5(\ln x + 1) + x^5$ .
Combine Like Terms: Factor out the common factor: $(\frac{dy}{dx}) = x^5(6\ln x + 7)$ .
Factor Out Common Factor: For the fourth function $y = e^{2x}(5x + 2)$ , let $f = e^{2x}$ and $g = 5x + 2$ .
Apply Product Rule: Differentiate $f = e^{2x}$ to get $f' = 2e^{2x}$ .
Differentiate $f$ : Differentiate $g = 5x + 2$ to get $g' = 5$ .
Differentiate $g$ : Apply the product rule: $\frac{dy}{dx} = f'g + fg' = 2e^{2x}(5x + 2) + e^{2x}(5)$ .
Apply Product Rule: Simplify the expression: $(\frac{dy}{dx}) = 10xe^{2x} + 4e^{2x} + 5e^{2x}$ .
Simplify Expression: Combine like terms: $(\frac{dy}{dx}) = 10xe^{(2x)} + 9e^{(2x)}.$
Combine Like Terms: Factor out the common factor: $(\frac{dy}{dx}) = e^{(2x)}(10x + 9)$ .
Factor Out Common Factor: For the fifth function $y = x^5e^{2x}$ , let $f = x^5$ and $g = e^{2x}$ .
Apply Product Rule: Differentiate $f = x^5$ to get $f' = 5x^4$ .
Differentiate $f$ : Differentiate $g = e^{2x}$ to get $g' = 2e^{2x}$ .
Differentiate $g$ : Apply the product rule: $\frac{dy}{dx} = f'g + fg' = 5x^4e^{2x} + x^5(2e^{2x})$ .
Apply Product Rule: Simplify the expression: $(\frac{dy}{dx}) = 5x^4e^{(2x)} + 2x^5e^{(2x)}$ .
Simplify Expression: Factor out the common factor: $(\frac{dy}{dx}) = x^{4}e^{(2x)}(5 + 2x)$ .
Factor Out Common Factor: For the sixth function $y = 5x(x^2 - 2)^3$ , let $f = 5x$ and $g = (x^2 - 2)^3$ .
Apply Product Rule: Differentiate $f = 5x$ to get $f' = 5$ .
Differentiate $f$ : Differentiate $g = (x^2 - 2)^3$ using the chain rule to get $g' = 3(x^2 - 2)^2(2x)$ .
Differentiate $g$ : Apply the product rule: $(\frac{dy}{dx}) = f'g + fg' = 5(x^2 - 2)^3 + 5x(3(x^2 - 2)^2(2x))$ .
Apply Product Rule: Simplify the expression: $(\frac{dy}{dx}) = 5(x^2 - 2)^3 + 30x^2(x^2 - 2)^2$ .
Simplify Expression: Factor out the common factor: $(\frac{dy}{dx}) = 5(x^2 - 2)^2((x^2 - 2) + 6x^2)$ .
Factor Out Common Factor: Simplify the expression inside the brackets: $(\frac{dy}{dx}) = 5(x^2 - 2)^2(7x^2 - 2)$ .