Help to find the equation of the tangent line to the curve y=x at the point (9,3).

Let f(x)=xaf′(x)=a×xa−1f(x)=x2, then f′(x)=2×x1f(x)=x3, f′(x)=3×x2Thus, if f9x)=x12, what is f'(x)?

I have this: f(x+h)−f(x)h=9+h+39+h+3×9+h−3h=9+h−3h9+h+3=69+h+3=66=1

Let f(x)=xaf′(x)=a×xa−1f(x)=x2, then f′(x)=2×x1f(x)=x3, f′(x)=3×x2Thus, if f9x)=x12, what is f'(x)?

I have this: f(x+h)−f(x)h=9+h+39+h+3×9+h−3h=9+h−3h9+h+3=69+h+3=66=1

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High School
Calculus and Analysis
Calculus 1
Derivatives

Stefan Hendricks

Answered question

2021-12-07

Help to find the equation of the tangent line to the curve

y = \sqrt{x}

at the point (9,3).

Answer & Explanation

Stella Calderon

Beginner2021-12-08Added 35 answers

Let $f (x) = x^{a}$
$f^{'} (x) = a \times x^{a - 1}$
$f (x) = x^{2}$ , then $f^{'} (x) = 2 \times x^{1}$
$f (x) = x^{3}$ , $f^{'} (x) = 3 \times x^{2}$
Thus, if $f 9 x) = x^{\frac{1}{2}}$ , what is f'(x)?

Jordan Mitchell

Beginner2021-12-09Added 31 answers

I have this:

\frac{f (x + h) - f (x)}{h} = \frac{\sqrt{9 + h} + 3}{\sqrt{9 + h} + 3} \times \frac{\sqrt{9 + h} - 3}{h} = \frac{9 + h - 3}{h \sqrt{9 + h} + 3} = \frac{6}{\sqrt{9 + h} + 3} = \frac{6}{6} = 1

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