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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Choose the correct answer in the slope of the tangent to the curve \(x = t^2 + 3t – 8, y = 2t^2 – 2t – 5\) at the point \((2,– 1)\) is

\[(A)\; \frac{22}{7} \quad (B)\; \frac{6}{7} \quad (C) \;\frac{7}{6} \quad (D) \;\frac{-6}{7}\]

Can you answer this question?

1 Answer

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  • $\large\frac{d}{dx}$$(x^n)=nx^{n-1}$
Step 1:
The curve is $x=t^2+3t-8,y=2t^2-2t-5$------(1)
Put $x=2$
Step 2:
Put $t=2$ in
At $x=2,y=-1,t=2$
Step 3:
Differentiating (1) with respect to $t$
$\large\frac{dy}{dx}=$$4t-2\times \large\frac{1}{2t+3}$
Step 4:
At $t=2$
$\large\frac{dy}{dx}=\frac{4\times 2-2}{2\times 2+3}$
Part (B) is the correct answer.
answered Aug 13, 2013 by sreemathi.v
edited Sep 2, 2013 by sharmaaparna1

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