i.1). $y=21 \cdot e^{x}+4 \cdot 2^{x}+5 ; \quad y^{\prime}(0)=$ ?
2). $y=\frac{\sqrt{x}-2}{x+4 x^{2}} ; y^{\prime}(1)=$ ?
3) $y=(1+\sin x) \cdot(2-\cos x) ; \quad y^{\prime}(0)=?$

Question

i.1). $y=21 \cdot e^{x}+4 \cdot 2^{x}+5 ; \quad y^{\prime}(0)=$ ?
2). $y=\frac{\sqrt{x}-2}{x+4 x^{2}} ; y^{\prime}(1)=$ ?
3) $y=(1+\sin x) \cdot(2-\cos x) ; \quad y^{\prime}(0)=?$

Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find $y'(0)$ for $y=21 \cdot e^{x}+4 \cdot 2^{x}+5$, we use the derivative $y'(x) = 21 e^x + 2^{x + 2} \log(2)$ provided by asksia-ll. ‖ ‖ step 2 ⋮ Substitute $x=0$ into the derivative: $y'(0) = 21 e^0 + 2^{0 + 2} \log(2)$. ‖ ‖ step 3 ⋮ Simplify the expression: $y'(0) = 21 \cdot 1 + 4 \cdot \log(2)$. ‖ ‖ step 4 ⋮ Calculate the value: $y'(0) = 21 + 4 \log(2)$. ‖ ∻Answer∻ ⚹ $y'(0) = 21 + 4 \log(2)$ ⚹ ∻Key Concept∻ ⚹Differentiation and evaluation at a point⚹ ∻Explanation∻ ⚹The derivative of the function at $x=0$ is found by substituting $x=0$ into the derivative and simplifying.⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To find $y'(1)$ for $y=\frac{\sqrt{x}-2}{x+4 x^{2}}$, we use the derivative result from asksia-ll. ‖ ‖ step 2 ⋮ Evaluate the derivative at $x=1$: $y'(1) = \frac{23}{50}$. ‖ ∻Answer∻ ⚹ $y'(1) = \frac{23}{50}$ ⚹ ∻Key Concept∻ ⚹Evaluation of a derivative at a specific point⚹ ∻Explanation∻ ⚹The derivative of the function at $x=1$ is directly given by the asksia-ll calculation.⚹ ∻Solution by Steps∻ ‖ step 1 ⋮ To find $y'(0)$ for $y=(1+\sin x) \cdot(2-\cos x)$, we use the derivative result from asksia-ll. ‖ ‖ step 2 ⋮ Evaluate the derivative at $x=0$: $y'(0) = 1$. ‖ ∻Answer∻ ⚹ $y'(0) = 1$ ⚹ ∻Key Concept∻ ⚹Product rule and evaluation of a derivative⚹ ∻Explanation∻ ⚹The derivative of the product of two functions at $x=0$ is found by applying the product rule and then evaluating at $x=0$.⚹◊What is the derivative of the function $y=21 \cdot e^{x}+4 \cdot 2^{x}+5$ evaluated at $x=0$?⍭ Generate me a similar question◊

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