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Continuity and Differentiability

Short Answer Type

491.

Differentiate space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction close parentheses space straight w. straight r. straight t. space sin to the power of negative 1 end exponent open parentheses fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction close parentheses.

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492.

Differentiate space tan to the power of negative 1 end exponent open parentheses fraction numerator 3 straight x minus straight x cubed over denominator 1 minus 3 straight x squared end fraction close parentheses space straight w. straight r. straight t. space tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator square root of 1 minus straight x squared end root end fraction close parentheses

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493. Prove that the derivative of tan^-1 $open parentheses fraction numerator square root of 1 plus straight x squared end root minus 1 over denominator straight x end fraction close parentheses$ w.r.t. tan^-1 x is independent of x.

Let space straight y equals tan to the power of negative 1 end exponent open parentheses fraction numerator square root of 1 plus straight x squared end root minus 1 over denominator straight x end fraction close parentheses space comma space straight u equals tan to the power of negative 1 end exponent straight x Put space straight x equals tan space straight theta therefore space space space straight y equals tan to the power of negative 1 end exponent open square brackets fraction numerator square root of 1 plus tan squared space straight theta end root minus 1 over denominator tan space straight theta end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator sec space straight theta minus 1 over denominator tan space straight theta end fraction close square brackets space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator begin display style fraction numerator 1 over denominator cos space straight theta end fraction end style minus 1 over denominator begin display style fraction numerator sin space straight theta over denominator cos space straight theta end fraction end style end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator 1 minus cos space straight theta over denominator sin space straight theta end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator 2 sin 2 begin display style straight theta over 2 end style over denominator 2 sin begin display style straight theta over 2 end style space cos begin display style straight theta over 2 end style end fraction close square brackets space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator sin begin display style straight theta over 2 end style over denominator cos begin display style straight theta over 2 end style end fraction close square brackets equals tan to the power of negative 1 end exponent open parentheses tan straight theta over 2 close parentheses equals straight theta over 2 therefore space straight y equals 1 half tan to the power of negative 1 end exponent straight x space rightwards double arrow space dy over dx equals fraction numerator 1 over denominator 2 left parenthesis 1 plus straight x squared right parenthesis end fraction Also space straight u equals tan to the power of negative 1 end exponent straight x space space rightwards double arrow space space dy over dx equals fraction numerator 1 over denominator 1 plus straight x squared end fraction Now space dy over dx equals fraction numerator begin display style dy over dx end style over denominator begin display style du over dx end style end fraction equals fraction numerator 1 over denominator 2 left parenthesis 1 plus straight x squared right parenthesis end fraction cross times fraction numerator 1 plus straight x to the power of 1 over denominator 1 end fraction equals 1 half which space does space not space contain space straight x space and space therefore space is space independent space of space straight x.

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494.

Find space the space derivative space of space tan to the power of negative 1 end exponent fraction numerator 2 straight x over denominator 1 minus straight x squared end fraction space straight w. straight r. straight t. space tan to the power of negative 1 end exponent straight x

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495.

Differentiate space sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 2 straight x squared minus 1 end fraction close parentheses space straight w. straight r. straight t. space sin to the power of negative 1 end exponent left parenthesis 3 straight x minus 4 straight x cubed right parenthesis.

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496.

Differentiate space sin to the power of negative 1 end exponent straight x space straight w. straight r. straight t. space cos to the power of negative 1 end exponent square root of 1 minus straight x squared end root.

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497.

Differentiate space sin to the power of negative 1 end exponent open parentheses fraction numerator 1 minus straight x over denominator 1 plus straight x end fraction close parentheses space straight w. straight r. straight t. space square root of straight x

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498.

Differentiate space cos to the power of negative 1 end exponent straight theta space straight w. straight r. straight t. space log left parenthesis 1 plus straight theta right parenthesis

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499.

Differentiate space sin to the power of negative 1 end exponent straight theta space straight w. straight r. straight t. space log left parenthesis 1 plus straight theta right parenthesis

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500.

Differentiate space log left parenthesis 1 plus straight theta right parenthesis space straight w. straight r. straight t. space sin to the power of negative 1 end exponent space straight theta

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Previous Year Papers

Test Series

Test Yourself

Short Answer Type

493. Prove that the derivative of tan-1 w.r.t. tan-1 x is independent of x.

493. Prove that the derivative of tan^-1 $open parentheses fraction numerator square root of 1 plus straight x squared end root minus 1 over denominator straight x end fraction close parentheses$ w.r.t. tan^-1 x is independent of x.