name: max points: 100 answer all the questions. each carries 20 points. 1. find d/dx(17x³ - 4e^x - 5ln(x) +…

1.

Explanation:

Step1: Apply sum - difference rule of derivatives

$\frac{d}{dx}(17x^{3}-4e^{x}-5\ln(x)+14\sqrt{x}-3)=\frac{d}{dx}(17x^{3})-\frac{d}{dx}(4e^{x})-\frac{d}{dx}(5\ln(x))+\frac{d}{dx}(14\sqrt{x})-\frac{d}{dx}(3)$

Step2: Use power rule for $17x^{3}$

$\frac{d}{dx}(17x^{3}) = 17\times3x^{2}=51x^{2}$

Step3: Use derivative of exponential for $4e^{x}$

$\frac{d}{dx}(4e^{x}) = 4e^{x}$

Step4: Use derivative of natural logarithm for $5\ln(x)$

$\frac{d}{dx}(5\ln(x))=\frac{5}{x}$

Step5: Rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ and use power rule

$\frac{d}{dx}(14\sqrt{x})=\frac{d}{dx}(14x^{\frac{1}{2}})=14\times\frac{1}{2}x^{-\frac{1}{2}} = \frac{7}{\sqrt{x}}$

Step6: Derivative of a constant

$\frac{d}{dx}(3) = 0$

Answer:

$51x^{2}-4e^{x}-\frac{5}{x}+\frac{7}{\sqrt{x}}$

2.

Explanation:

Step1: Find the derivative of $f(x)$

$f(x)=10 - x^{2}$, so $f'(x)=-2x$

Step2: Evaluate the derivative at $x = 2$

$f'(2)=-2\times2=-4$, this is the slope $m$ of the tangent line

Step3: Find the $y$-coordinate of the point on the curve at $x = 2$

$f(2)=10-2^{2}=6$, so the point $(x_0,y_0)=(2,6)$

Step4: Use the point - slope form of a line $y - y_0=m(x - x_0)$

$y - 6=-4(x - 2)$

Step5: Simplify the equation

$y-6=-4x + 8$, so $y=-4x+14$

Answer:

$y=-4x + 14$

3.

Explanation:

Step1: Recall the product rule $(uv)'=u'v+uv'$

Let $u = 8x^{5}-9x^{4}$ and $v=e^{2x}+5$

Step2: Find $u'$

$u'=\frac{d}{dx}(8x^{5}-9x^{4})=8\times5x^{4}-9\times4x^{3}=40x^{4}-36x^{3}$

Step3: Find $v'$

$v'=\frac{d}{dx}(e^{2x}+5)=2e^{2x}$

Step4: Apply the product rule

$h'(x)=(40x^{4}-36x^{3})(e^{2x}+5)+(8x^{5}-9x^{4})\times2e^{2x}$

Answer:

$(40x^{4}-36x^{3})(e^{2x}+5)+(8x^{5}-9x^{4})\times2e^{2x}$